Proving that a space in $\ell_{\infty}$ is closed. An exercise in my notes is to prove that the space 
$$
Y = \left\{\left(y_n\right)_{n=1}^{\infty} \in \ell_{\infty} : \lim_{n \to \infty} y_{n} = 0\right\}
$$
is complete, by showing that it is first closed. 
To me this seemed rather not so hard, and so I wanted to make sure I have the right idea (or am not missing a more simple idea). Furthermore, using the fact that $F$ is closed seemed a bit needless to me as I will show. I apologize if this question seems far too simple for here. 
My proof. 
Clearly the sequence $\left(a_{n}\right)_{n} = \left(0, 0, 0, \dots\right) \in Y$, and so $Y$ contains all of its limit points as its only limit point is trivially $0$. Thus, $Y$ is closed in $\ell_{\infty}$. Then as we already know that $(\ell_{\infty}, d_{\infty})$ is a complete metric space and $Y \subseteq \ell_{\infty}$ is closed, it follows that $Y$ is complete, as $\left(y_{n}\right)_{n}$ being Cauchy in $Y$ implies it is convergent in $\ell_{\infty}$, but since $Y$ is closed $\left(y_{n}\right)_{n}$ must then converge in $Y$. $\square$
My follow up question:
Is it really necessary to use the fact that $Y$ is closed here? My first thought would simply be to show that every Cauchy sequence in $Y$ converges to $0 \in Y$ (just using $\triangle$ inequality) and conclude completeness directly from the definition. Would there be anything wrong with that? This particular question has seemed simple enough I fear I have done something very wrong. 
 A: I see with @LeBtz's help that my attempted proof above had a glaring issue, and I was rather confused with the notion of convergence and Cauchy-ness in $Y$. In particular, $Y$ does not only contain $0$ as a limit point.  I think however that I have constructed a correct proof now, by showing that $\ell_{\infty} \setminus Y$ is open. (If there's still any issues please let me know!)

Consider some $(y_{n})_{n} \in \ell_{\infty} \setminus Y$. Then $(y_{n})_{n}$ cannot be convergent to $0$, and so for some $\frac{\epsilon}{2} > 0$ we can find infinitely many indices $n \in \mathbb{N}$ such that $d_{\infty}(y_{n}, 0) \geq \epsilon$. Denote this set of indices by $N$. Then for $(z_{n})_{n} \in B\left(\left(y_{n}\right)_{n}, \frac{\epsilon}{2}\right)$, we have that $d_{\infty}(z_{n}, 0) \geq \frac{\epsilon}{2} \ \forall n \in N$, and hence $B\left(\left(y_{n}\right)_{n}, \frac{\epsilon}{2}\right) \subseteq \ell_{\infty} \setminus Y$. 
That is, $\ell_{\infty} \setminus Y$ is open, and so $Y$ is closed. Now it follows as shown in the incorrect proof above that $Y$ is complete. 
A: You are confused with the definition of a closed space. You have to take a convergent sequence over elements of the space, and every vector of that sequence consists of infinitely many elements.
Let $\{y^{(j)}\}_{j\in\mathbb{N}}$ a sequence of elements in $Y$, that converges to $y\in\ell_\infty$. Let $n\in\mathbb{N}$. Then, as $y^{(j)}\to y$ in the $||\cdot||_\infty$ norm, we have that there is a natural $J_n$ such that for all $j>J_n$: $$||y-y^{(j)}||_\infty<1/n$$ Now, see that for all fixed $n$, if $j>J_n$ we have: $$|y_n|\leq|y_n-y_n^{(j)}| + |y_n^{(j)}|\leq||y-y^{(j)}||_\infty + |y_n^{(j)}| < \frac{1}{n} + |y_n^{(j)}|$$ And now just taking limits with $n\to\infty$, as $y^{(j)}\in Y$, we conclude that $$\lim_{n\to\infty}y_n=0$$ So $y\in Y$ and then $Y$ is closed.
