Let's prove the theorem for real-valued functions.
First: notations and useful results
If $f:\mathbb{R}\to\mathbb{R}$ is a $2\pi$-periodic function continuous from the right of bounded variation over a period, denote the Lebesgue-Stieltjes signed measure associated to $f$ with $\mu_f$, i.e. $\mu_f$ is the only signed measure such that
$$\mu _f((a,b])=f(b)-f(a).$$
Also, define $V_f:\mathbb{R}\to \mathbb{R}$ as a variation of $f$, i.e.
$$|\mu_f|((a,b])=V_f(b)-V_f(a).$$
Recall that if $\varphi\in C_c^1(\mathbb{R})$ then it holds the follow integration by parts formula:
$$\int_\mathbb{R}f(t)\varphi'(t)\operatorname{d}t=-\int_\mathbb{R}\varphi(t)\operatorname{d}\mu_f(t)$$
that leads to:
$$\forall\varphi\in C^1(\mathbb{R}), \forall a\in\mathbb{R}, \forall b>a, \int_a^bf(t)\varphi'(t)\operatorname{d}t=f(b^+)\varphi(b)-f(a^-)\varphi(a)-\int_{[a,b]}\varphi(t)\operatorname{d}\mu_f(t).$$
In our case, $f$ is also continuous, so actually we also have that:
$$\mu _f([a,b])=f(b)-f(a) \\ |\mu _f|([a,b])=V_f(b)-V_f(a)\\ \int_a^bf(t)\varphi'(t)\operatorname{d}t=f(b)\varphi(b)-f(a)\varphi(a)-\int_{[a,b]}\varphi(t)\operatorname{d}\mu_f(t).$$
Also, define $f_x(t):=f(x+t)-f(x)$ and notice that $\mu_{f_x}(A)=\mu_{f}(x+A)$, $|\mu_{f_x}|(A)=|\mu_{f}|(x+A)$ and $V_{f_x}(t)=V_f(x+t).$
Also, define $$\varphi_N(s):=\int_0^s \frac{\sin((N+\frac{1}{2})t)}{\sin(\frac{t}{2})}\operatorname{d}t$$
In the answer to this question it is proved that:
$$\exists C>0, \forall s\in[-\pi,\pi], \forall N\in\mathbb{N}, |\varphi_N(s)|\le C.$$
Second: a reformulation of what we want to prove
We have that:
$$\sum_{n=-N}^N \hat{f}(n)e^{inx}-f(x)=\int_{-\pi}^\pi(f(x+t)-f(x))\frac{\sin((N+\frac{1}{2})t)}{\sin(\frac{t}{2})}\frac{\operatorname{d}t}{2\pi} = \int_{-\pi}^\pi f_x(t)\frac{\sin((N+\frac{1}{2})t)}{\sin(\frac{t}{2})}\frac{\operatorname{d}t}{2\pi},$$
so we want to prove that:
$$\sup_{x\in[-\pi,\pi]}\left|\int_{-\pi}^\pi f_x(t)\frac{\sin((N+\frac{1}{2})t)}{\sin(\frac{t}{2})}\frac{\operatorname{d}t}{2\pi}\right|\to 0, N\to +\infty.$$
Now:
If $\delta\in(0,\pi)$ we have that:
$$\sup_{x\in[-\pi,\pi]}\left|\int_{-\pi}^\pi f_x(t)\frac{\sin((N+\frac{1}{2})t)}{\sin(\frac{t}{2})}\frac{\operatorname{d}t}{2\pi}\right|\\ \le \sup_{x\in[-\pi,\pi]}\left|\int_{-\delta}^\delta f_x(t)\frac{\sin((N+\frac{1}{2})t)}{\sin(\frac{t}{2})}\frac{\operatorname{d}t}{2\pi}\right| + \sup_{x\in[-\pi,\pi]}\left|\int_{[-\pi,\pi]\backslash [-\delta,\delta]} f_x(t)\frac{\sin((N+\frac{1}{2})t)}{\sin(\frac{t}{2})}\frac{\operatorname{d}t}{2\pi}\right|.$$
So it enough to prove that for all $\varepsilon>0$ there exists $\delta\in(0,\pi)$ such that:
$$\forall N\in\mathbb{N}, \sup_{x\in[-\pi,\pi]}\left|\int_{-\delta}^\delta f_x(t)\frac{\sin((N+\frac{1}{2})t)}{\sin(\frac{t}{2})}\frac{\operatorname{d}t}{2\pi}\right|\le \frac{C}{\pi}\varepsilon$$
and:
$$\sup_{x\in[-\pi,\pi]}\left|\int_{[-\pi,\pi]\backslash [-\delta,\delta]} f_x(t)\frac{\sin((N+\frac{1}{2})t)}{\sin(\frac{t}{2})}\frac{\operatorname{d}t}{2\pi}\right|\to0, N\to+\infty.$$
Third: first integral estimate
Let's use the integration by parts formula:
$$\left|\int_{-\delta}^\delta f_x(t)\frac{\sin((N+\frac{1}{2})t)}{\sin(\frac{t}{2})}\frac{\operatorname{d}t}{2\pi}\right|=\left|\int_{-\delta}^\delta f_x(t) \varphi_N'(t)\frac{\operatorname{d}t}{2\pi}\right|\\ = \frac{1}{2\pi}\left|-\int_{[-\delta,\delta]} \varphi_N(t)\operatorname{d}\mu_{f_x}(t)+f_x(\delta)\varphi_N(\delta)-f_x(-\delta)\varphi_N(-\delta)\right|\le \frac{C}{\pi} |\mu_{f_x}|([-\delta,\delta]) \\ = \frac{C}{\pi} |\mu_{f}|([x-\delta,x+\delta]) = \frac{C}{\pi} (V_f(x+\delta)-V_f(x+\delta)).$$
Now, being $f$ continuous, we have that $V_f$ is continuous and so uniformly continuous e.g. over the interval $[-2\pi,2\pi]$. So, if $\varepsilon>0$ and $\delta\in(0,\frac{\pi}{2})$ is such that for $|x-y|\le 2\delta$ we have that $|V_f(x)-V_f(y)|<\varepsilon$, then we have that:
$$\forall x\in [-\pi,\pi], (V_f(x+\delta)-V_f(x+\delta)) <\varepsilon$$
and so:
$$\forall N\in\mathbb{N}, \sup_{x\in[-\pi,\pi]}\left|\int_{-\delta}^\delta f_x(t)\frac{\sin((N+\frac{1}{2})t)}{\sin(\frac{t}{2})}\frac{\operatorname{d}t}{2\pi}\right|\le \frac{C}{\pi}\varepsilon$$
Fourth: second integral estimate
Since:
$$\sin\left((N+\frac{1}{2})t\right)= \sin (Nt) \cos(\frac{t}{2})+\cos (Nt) \sin(\frac{t}{2}),$$ we have that:
$$\frac{\sin((N+\frac{1}{2})t)}{\sin(\frac{t}{2})}=\frac{\sin(Nt)}{\tan(\frac{t}{2})}+\cos(Nt).$$
Now:
$$\int_{-\pi}^{\pi} f_x(t) \cos(Nt) \operatorname{d}t = -\int_{-\pi}^{\pi} f_x(t-\frac{\pi}{N}) \cos(Nt) \operatorname{d}t,$$
so:
$$\left|\int_{-\pi}^{\pi} f_x(t) \cos(Nt) \operatorname{d}t\right| = \left|\frac{1}{2} \int_{-\pi}^{\pi} (f_x(t)-f_x(t-\frac{\pi}{N})) \cos(Nt) \operatorname{d}t\right| = \left| \frac{1}{2} \int_{-\pi}^{\pi} (f(t)-f(t-\frac{\pi}{N})) \cos(Nt) \operatorname{d}t\right|\le \frac{1}{2} \int_{-\pi}^{\pi} \left|(f(t)-f(t-\frac{\pi}{N}))\right| \operatorname{d}t\le \omega_{f,1}(\frac{\pi}{N}) $$
where $$\omega_{g,1}(\alpha)=\sup_{h\in[-\alpha,\alpha]}\int_{-\pi}^{\pi} |g(t+h)-g(t)|\operatorname{d}t$$
and then:
$$\sup_{x\in[-\pi,\pi]}\left|\int_{[-\pi,\pi]\backslash [-\delta,\delta]} f_x(t)\cos(Nt)\frac{\operatorname{d}t}{2\pi}\right|\le \sup_{x\in[-\pi,\pi]}\left|\int_{[-\pi,\pi]} f_x(t)\cos(Nt)\frac{\operatorname{d}t}{2\pi}\right|
\le \frac{1}{2\pi} \omega_{f,1}(\frac{\pi}{N})$$
So, since $f\in L^1([-\pi,\pi])$, we have that
$$\omega_{f,1}(\frac{\pi}{N})\to 0, N\to+\infty$$
and then:
$$\sup_{x\in[-\pi,\pi]}\left|\int_{[-\pi,\pi]\backslash [-\delta,\delta]} f_x(t)\cos(Nt)\frac{\operatorname{d}t}{2\pi}\right|\to 0, N\to+\infty.$$
So, it remains to prove that:
$$\sup_{x\in[-\pi,\pi]}\left|\int_{[-\pi,\pi]\backslash [-\delta,\delta]} f_x(t)\frac{\sin(Nt)}{\tan(\frac{t}{2})}\frac{\operatorname{d}t}{2\pi}\right|\to 0, N\to+\infty.$$
Now, let $\psi$ be a continuous $2\pi$-function that coincides with $t\mapsto \frac{1}{\tan(\frac{t}{2})}$ on $[-\pi,\pi]\backslash[-\delta,\delta]$. Then, what we want to prove is equivalent to:
$$\sup_{x\in[-\pi,\pi]}\left|\int_{[-\pi,\pi]\backslash [-\delta,\delta]} f_x(t)\psi(t)\sin(Nt)\frac{\operatorname{d}t}{2\pi}\right|\to 0, N\to+\infty.$$
With the same technique as before, we have that:
$$\left|\int_{[-\pi,\pi]} f_x(t)\psi(t)\sin(Nt){\operatorname{d}t}\right|\le \left|\frac{1}{2} \int_{-\pi}^{\pi} (f_x(t)\psi(t)-f_x(t-\frac{\pi}{N}))\psi(t-\frac{\pi}{N})) \cos(Nt) \operatorname{d}t\right|\\ \le \frac{1}{2} \int_{-\pi}^{\pi}\left| f_x(t)\psi(t)-f_x(t-\frac{\pi}{N})\psi(t-\frac{\pi}{N}))\right|\operatorname{d}t \le \frac{1}{2} \int_{-\pi}^{\pi}\left| f_x(t)\psi(t)-f_x(t-\frac{\pi}{N})\psi(t)+f_x(t-\frac{\pi}{N})\psi(t)-f_x(t-\frac{\pi}{N})\psi(t-\frac{\pi}{N}))\right|\operatorname{d}t \\ \le \frac{1}{2} \int_{-\pi}^{\pi}\left| f_x(t)\psi(t)-f_x(t-\frac{\pi}{N})\psi(t)\right|\operatorname{d}t+ \frac{1}{2} \int_{-\pi}^{\pi}\left|f_x(t-\frac{\pi}{N})\psi(t)-f_x(t-\frac{\pi}{N})\psi(t-\frac{\pi}{N}))\right|\operatorname{d}t \\ \le \frac{\|\psi\|_\infty}{2} \int_{-\pi}^{\pi}\left| f_x(t)-f_x(t-\frac{\pi}{N})\right|\operatorname{d}t+ \frac{\|f\|_\infty}{2} \int_{-\pi}^{\pi}\left|\psi(t)-\psi(t-\frac{\pi}{N}))\right|\operatorname{d}t \\ \le \frac{\|\psi\|_\infty}{2} \omega_{f,1}(\frac{\pi}{N})+ \frac{\|f\|_\infty}{2} \omega_{\psi,1}(\frac{\pi}{N})$$
and since $f,\psi\in L^1([-\pi,\pi])$
$$\sup_{x\in[-\pi,\pi]}\left|\int_{[-\pi,\pi]\backslash [-\delta,\delta]} f_x(t)\psi(t)\sin(Nt)\frac{\operatorname{d}t}{2\pi}\right|\le\frac{1}{2\pi}\left(\frac{\|\psi\|_\infty}{2} \omega_{f,1}(\frac{\pi}{N})+ \frac{\|f\|_\infty}{2} \omega_{\psi,1}(\frac{\pi}{N})\right)\to 0, N\to+\infty.$$
