I wrote a proof and would be happy for your feedback since this exercise gives me extra points in the exam.
$\textbf{Proof.}$ Let $f: I \to \mathbb{R}$ be a convex function, where $I \subseteq \mathbb{R}$ is any non-trivial interval. Choose $a, b \in I$ with $a < b$ and $t \in (0,1)$. We aim to show that $f$ is left and right differentiable at $x \in I$. Assume $x \in I$ such that $a < x < b$.
Since $f$ is convex, for all $a, b \in I$ with $a < b$, the convexity inequality holds: $$\[f((1-t)a + tb) \leq (1-t)f(a) + tf(b) \quad \forall t \in (0,1).\]$$
Let $t = \frac{x-a}{b-a}$. Since $a < x < b$, it follows that $0 < x-a < b-a$, implying $0 < \frac{x-a}{b-a} < 1$. Thus, our chosen $t$ is in $(0,1)$. Substituting $t$ into the convexity inequality and multiplying by $-xf(x)$, we get: $$\begin{align*} f(x) &\leq \left(1 - \frac{x-a}{b-a}\right)f(a) + \frac{x-a}{b-a}f(b) \\ f(x) &\leq \left(\frac{b-x}{b-a}\right)f(a) + \frac{x-a}{b-a}f(b) \\ (b-a)f(x) &\leq (b-x)f(a) + (x-a)f(b) \\ bf(x) - af(x) - xf(x) &\leq bf(a) -xf(a) +xf(b) -af(b) -xf(x) \\ \left( f(x) -f(a)\right)(b-x) &\leq (f(b)-f(x))(x-a) \\ \frac{f(x)-f(a)}{x-a} &\leq \frac{f(b)-f(x)}{b-x}. \end{align*}$$
We observe that $\frac{f(x)-f(a)}{x-a} = \frac{-(f(a)-f(x))}{-(a-x)} = \frac{f(a)-f(x)}{a-x}.$
Now, define the function $g:[a,b]\setminus\{x\} \to \mathbb{R},\; z \mapsto \frac{f(z)-f(x)}{z-x}.$ Letting the function approach $x$ from the left for $a \in [a,x)$ and from the right for $b \in (x,b]$, we obtain: $$\[f_-^\prime(x)= \lim_{a \to x^-}\frac{f(a)-f(x)}{(a-x)} \quad \text{and} \quad \lim_{b \to x^+}\frac{f(b)-f(x)}{(b-x)} = f_+^\prime(x).\]$$
In particular, with the above inequality: $$\[f_-^\prime(x) \leq \lim_{a \to x^-}\frac{f(a)-f(x)}{(a-x)} \leq \lim_{b \to x^+}\frac{f(b)-f(x)}{(b-x)} = f_+^\prime(x),\]$$ where left and right limits may not necessarily coincide.
As $x$ was chosen arbitrarily, the statement follows. $\Box$
$\textbf{Proof.}$ We want to show that a convex function $f:I \to \mathbb{R}$ is continuous. Let $x_0 \in I$ and let $\varepsilon > 0$ be arbitrary. Also, let $\delta > 0$ be chosen such that for $x \in (x_0 - \delta, x_0)$ and $x \in (x_0, x_0 + \delta)$, the argumentation regarding the left and right-sided derivatives is correct.
For $x \in (x_0 - \delta, x_0)$, due to the existence of the left-sided derivative: $$\begin{aligned} f_-^\prime(x_0) &= \lim_{\substack{x \to x_0\\ x < x_0}}\frac{f(x)-f(x_0)}{x-x_0} \\ \implies \lim_{\substack{x \to x_0\\ x < x_0}}(x-x_0)\cdot f^\prime_- (x_0) &= \lim_{\substack{x \to x_0\\ x < x_0}}f(x)-f(x_0) \\ \implies \lim_{\substack{x \to x_0\\ x < x_0}}(x-x_0)\cdot f^\prime_- (x_0) + f(x_0) &= \lim_{\substack{x \to x_0\\ x < x_0}}f(x) \\ \implies \varepsilon+f(x_0) &= \lim_{\substack{x \to x_0\\ x < x_0}} f(x), \end{aligned}$$ which gives us $|f(x) - f(x_0)| < \varepsilon \;\; \forall x \in (x_0 - \delta, x_0)$.
The same argumentation follows for $x \in (x_0, x_0 + \delta)$ and the right-sided derivative. Since $x_0$ and $\varepsilon$ were arbitrary, continuity follows for the entire interval $I.$ $\Box$
