This is a proof I came up with while working on the textbook Understanding Analysis:
Supposing $x_{n} \rightarrow x$, we have that
$$s_n = \frac{1}{n} \sum_{k=1}^{n} x_k \rightarrow x$$
Let $\epsilon \in \mathbb{R}^{+}$
Notice that by the triangle equality we get: $$\vert \frac{1}{n} \sum_{k=1}^{n} x_k - x \vert = \vert \frac{1}{n} \sum_{k=1}^{n} (x_k - x) \vert \leq \frac{1}{n} \sum_{k=1}^{n} \vert x_k - x \vert$$
Now, since $x_{n} \rightarrow x$, we know that for each $n>N$ term, we can make $\vert x_n - x \vert$ as small as we wish. Let $M = \max(x_{1},...,x_{N})$, and let $ \vert x_n - x \vert < \vert \frac{\epsilon n - MN}{n-N}\vert$. Then assuming $\epsilon n - MN > 0$ we get that
$$\vert \frac{1}{n} \sum_{k=1}^{n} x_k - x \vert \leq \frac{1}{n} \sum_{k=1}^{n} \vert x_k - x \vert < \frac{NM + (n-N)\vert \frac{\epsilon n - MN}{n-N}\vert}{n} = \frac{MN + \epsilon n - MN}{n} = \epsilon$$
Notice, finally, that the assumption that $\epsilon n - MN > 0$ is harmless. This is since we let $n>\frac{MN}{\epsilon}$, after which the rest of the proof falls into place.