This question is asking for a proof of a claim I don't understand (namely the pointwise convergence of the partial sums to a Fourier series). Alternatively, a completely different approach that's not too complicated is also fine (i.e. it doesn't use things like the Dirichlet kernel or the Lebesgue dominated convergence theorem or fatou's lemma, etc.).
Let $f : [-\pi, \pi] \to\mathbb{C}$ be continuous be given by $f(x) = 1-|x/\pi|$ for each $x\in [-\pi, \pi]$. Let $A_n = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x)e^{-inx}dx$. Show that $\sum_{|n|\leq N} A_n e^{in\theta}$ converges uniformly to $f$.
I know how to show that $S_n(\theta) := \sum_{|n|\leq N}a_ne^{in\theta}$ converges uniformly, and this is by showing that $\{S_n\}_{n\ge 1}$ is Cauchy under the supremum metric. Since each $S_n(\theta) \in C([-\pi, \pi],\mathbb{C})$ (i.e. is a continuous function from $[-\pi,\pi]$ to $\mathbb{C}$), it must converge to some $g \in C([-\pi, \pi],\mathbb{C})$ by the completeness of the metric space $(C([-\pi, \pi],\mathbb{C}), \lVert \cdot \rVert_\infty)$.
If I could just show that $S_n$ converges pointwise to $f$, then since $S_n$ converges uniformly to $g$, it converges pointwise to $g$, and since pointwise limits are unique, this would show that $f = g$.
More details are shown below. Feel free to point out any errors I could have made.
First, we compute the Fourier coefficients of $f'$ using integration by parts. Suppose the Fourier coefficients of $f$ are $a_n :=\frac1\pi \int_{-\pi}^{\pi} f(x)\cos (nx)dx$ and $b_n := \frac1\pi\int_{-\pi}^{\pi}f(x)\sin (nx)dx$. Then $\frac1\pi\int_{-\pi}^\pi f'(x)\cos (nx)dx = \frac1\pi([f(x)\cos (nx)]_{-\pi}^\pi + n\int_{-\pi}^\pi f(x)\sin (nx)dx) = nb_n$ and $\frac1\pi\int_{-\pi}^{\pi} f'(x)\sin (nx)dx = \frac1\pi([f(x)\sin (nx)]_{-\pi}^\pi -n\int_{-\pi}^\pi f(x)\cos(nx)dx) = -na_n.$ So the Fourier series of $f'$ is $\sum_{n\ge 0} (nb_n \cos (nx) - na_n\sin (nx)).$
Note that the Fourier coefficients of $f'$ are square-summable (this follows for arbitrary functions from Bessel's inequality). That is, if $a_n = \frac{1}{\pi}\int_{-\pi}^{\pi} f'(x)\cos nx dx = \frac{1}{\pi} \langle f'(x), \cos nx\rangle$ and $b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f'(x)\sin nx dx= \frac{1}{\pi} \langle f'(x), \cos nx\rangle$, then $\sum_{n\ge 0} |a_n|^2 + |b_n|^2 \leq \frac{1}{\pi} \lVert f'\rVert_2^2 < \infty$, where $\langle f, g\rangle := \int_{-\pi}^{\pi} f(x)\overline{g(x)}dx$ for complex valued functions f and g and $\lVert f\rVert_2 := \sqrt{\langle f, f\rangle}$ for a function f.
By the Cauchy Schwarz inequality, we have that $(\sum_{n\ge 1} |n\cdot a_n\cdot \frac{1}n|)^2\leq \sum_{n\ge 1} (n^2 a_n^2)\cdot \sum_{n\ge 1} \frac{1}{n^2}<\infty,$ so $\sum_{n\ge 1} |a_n| < \infty.$ Similarly, $\sum_{n\ge 1} |b_n| < \infty$.
Note that if $\frac{c_0}2 + \sum_{n\ge 1}(c_n \cos (nx) + d_n \sin(nx)),d_0 := 0,$ is the Fourier series of a function $f$ and $\sum_{n\ge 1} (|c_n|+|d_n|) <\infty$, then by the Weierstrass M-test, $\sum_{|n|\leq N} (c_n \cos (nx) + d_n \sin(nx))\to \sum_{n\in\mathbb{Z}} (c_n \cos (nx) + d_n \sin(nx))$ uniformly. This can be applied to $f'$ by setting $c_n = a_n$ and $d_n=b_n$ for all n.
