My Proof: Every convergent sequence is a Cauchy sequence.

Idea is right, but the execution misses out on a couple of points.

So both will hold for all $n_1, n_2 > max(N_1, N_2)=N$, say $\epsilon = max(\epsilon_1, \epsilon_2)$

Technically $\,\epsilon\,$ is a given, you don't get to choose it.

then $\quad|x_{n_1}-x-(x_{n_2}-x)|<\epsilon \quad\implies\quad |x_{n_1}-x_{n_2}|<\epsilon$

The RHS does not follow from the stated premise that $\,|x_{n_1}-x| \lt \epsilon_1\,$ and $\,|x_{n_2}-x| \lt \epsilon_2$. At best, from the triangle inequality:

$$ |x_{n_1} - x_{n_2}| = |(x_{n_1}-x)-(x_{n_2}-x)| \le |x_{n_1}-x| + |x_{n_2}-x| \lt \epsilon_1 + \epsilon_2 $$

To fix it, just assume $\,\epsilon\,$ is given, choose $\,\epsilon_1=\epsilon_2=\epsilon / 2\,$, then proceed along the same line.

It should not be that for some $\epsilon_{1},\epsilon_{2}>0$. Rather, one fixes an arbitrary $\epsilon>0$, and we find $N_{1},N_{2}$ such that $|x_{n_{1}}-x|<\epsilon/2$ and $|x_{n_{2}}-x|<\epsilon/2$ for all $n_{1}>N_{1}$, $n_{2}>N_{2}$.

For all $n_{1},n_{2}>\max(N_{1},N_{2})$, then $|x_{n_{1}}-x_{n_{2}}|=|x_{n_{1}}-x-(x_{n_{2}}-x)|\leq|x_{n_{1}}-x|+|x_{n_{2}}-x|<\epsilon/2+\epsilon/2=\epsilon$.

Actually just one $N$ for which $|x_{n}-x|<\epsilon/2$, $n\geq N$ is enough.