Prove that $(l^\infty,\|.\|_\infty)$ is a Banach space.

Your proof is very rigorous and very detailed all the way up to the point where you say

Fix $m>n_0$. Then we have $\|x^m-x^n\|_\infty<\epsilon.$ Therefore $\|x^m-y\|_\infty<\epsilon$ as $n\to\infty$.

Now I know the inequality stands, but as you were very thorough with all your other inequalities, I think it would be nice if you wrote a little more justification for this one as well - it is not entirely obvious how the right inequality follows from the left one.

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Other than that, the proof is very well written and easy to follow.

Rather a comment, but my reputation does not allow me to comment yet. You write

choose $M>0$ that for each $n\in\mathbb{N}, |x^n_k|<M$

I am curios whether you have to write $M_k$, since hypothetically $M$ can depend on the choice of the sequence $(x^0_k,x^1_k,\cdots,x^n_k,\cdots)$. But then we have difficulties with proving that $y\in l^\infty$.

EDIT: the proof is more subtle than I initially thought. First we have to settle the issue 5xum mentioned. We start with $$ \forall k\in\mathbb{N}\ \forall \epsilon > 0 \ \exists n_0:\forall n,p>n_0:|x^n_k-x^{n+p}_k|<\epsilon. $$ This is inequality in $\mathbb{F}$, so we can let $p\to\infty$. Thus we obtain $$ \forall k\in\mathbb{N}\ \forall \epsilon > 0 \ \exists n_0:\forall n>n_0:|x^n_k-y_k|<\epsilon. $$ That means $\|x^n-y\|_\infty<\epsilon$, thus $(x^n)_{n\in\mathbb{N}}$ is converging to $y$.

We still have to prove that $y\in l^\infty$. But $$ \|y\|_\infty \le \|x^n-y\|_\infty + \|x^n\|_\infty \le \epsilon + \|x^n\|_\infty < \infty $$