Show that $C_{0} = \{(a_n)_{n \in \mathbb{N}}\subset \mathbb{R}: a_n \rightarrow 0\}$ is complete.
I've already seen that this question has been asked, and already answered, however, I've managed to make a different proof and I wanted to check it was ok, since yesterday I had some trouble when trying to prove something diferent which involved a similar step.
Ok, so $C_0 \subset \mathcal{l^{\infty}}$ which I already proved its complete. Therefore, its enough to check that $C_{0}$ is closed. So, given $(x^{(n)}_{k})_{n \in \mathbb{N}}$ where for every $n\in \mathbb{N}$, $x^{(n)}_{k}=(x_{1}^{(n)},x_{2}^{(n)},...,x_{k}^{(n)},...)$. Lets say $(x^{(n)}_{k})\rightarrow (a_k)$ as $n\rightarrow \infty$. I want to see that $(a_k)\in C_0$ To conclude that $C_0$ is closed.
That is, given $\epsilon > 0$ I want to find $k_0 \in \mathbb{N}$ such that $|a_k|<\epsilon$ if $k \geq k_0$.
Now, since $(x^{(n)}_{k})\rightarrow (a_k)$, given $\frac{\epsilon}{2}>0$ there exists $n_1 \in \mathbb{N}$ such that $d_{\infty}((x^{(n)}_{k}),(a_k)) = sup_{k \in \mathbb{N}}|x^{(n)}_{k} - a_k|<\frac{\epsilon}{2}$ if $n \geq n_1$.
Now, fixing $n\geq n_1$, we have that given $\frac{\epsilon}{2}>0$ there exists $k_0 \in \mathbb{N}$ such that $|x^{(n)}_{k}|<\frac{\epsilon}{2}$ if $k\geq k_0$. Therefore we can say that
$|a_k|-|x^{(n)}_{k}|\leq |a_k-x^{(n)}_{k}| \leq sup_{k \in \mathbb{N}}|x^{(n)}_{k} - a_k|<\frac{\epsilon}{2}$.
Taking $k\geq k_0$
$|a_k|<\frac{\epsilon}{2}+|x^{(n)}_{k}|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$
Concluding that $a_k \rightarrow 0$ and therefore, $a_k \in C_0$.
My only doubt about this proof is that my $k_0$ depends on the fixed $n\geq n_1$. But I don't see how that matters. (Im not sure if it does)!
Your proof is perfectly fine, moreover, it's very similar to the usual proof of that fact. The dependence exists, but it doesn't affect the proof (look at $n$ as an additional parameter that you use to get a value smaller, and once you've done that, it's not needed anymore).
I'll make a suggestion: don't say "given $\frac{\epsilon}{2}>0$", I'd say instead "for $\frac{\epsilon}{2}>0$" or "since $\frac{\epsilon}{2}>0$".