Proving that $Y= \{x_n \in \ell^2 : x_{2n} =0, n \in \mathbb N\}$ is closed and finding $Y^\bot$. Exercise :

Show that the space 
  $$Y= \{x_n \in \ell^2 : x_{2n} =0, n \in \mathbb N\}$$
  is a closed subspace of $\ell^2$ and find the space $Y^\bot$.

Attempt-Thoughts :
First of all, the $\ell^2$ space is defined as :
$$\ell^2 = \Big\{x=(x_n) : \sum_{i=1}^nx^2_n<+ \infty\Big\} \; \; \text{with the norm} \; \; \|x\|_2 = \sqrt{\sum_{i=1}^\infty x^2_n}$$
Of course, $\ell^2$ is not only a Banach space, but also a Hilbert space equipped with the inner product : $$\langle x,y\rangle=\sum_\limits{i=1}^\infty x_iy_i$$
For proving that $Y$ is a closed subspace of $\ell^2$, I know that if $H$ is a Hilbert space and $Y \subseteq H$ then $Y \subseteq Y^{\bot\bot}$ and $Y=Y^{\bot\bot}$ if and only if $Y$ is a closed subspace of $H$.
Finally, the orthogonal complement of $Y \subseteq\ell^2$, is defined as :
$$Y^\bot = \{x \in \ell^2 : \langle x,y\rangle = 0, \forall y \in Y\}$$
I assume that finding $Y^\bot$ follows imediatelly by the fact that since $Y$ is a closed subspace of $\ell^2$ which is a Hilbert space, then $\ell^2$ can be written as :
$$\ell^2 = Y \oplus Y^\bot$$
Question - Request : Any help, tips, hints or thorough elaborations regarding the exercise and mostly the case of $Y$ being a closed subspace of $\ell^2$ which I do now know how to handle, would be very much appreciated !

For each $k\in\mathbb N$, define$$\begin{array}{rccc}\pi_k\colon&\ell^2&\longrightarrow&\mathbb C\\&(x_n)_{n\in\mathbb N}&\mapsto&x_k.\end{array}$$Each $\pi_k$ is continuous and$$Y=\bigcap_{k\in\mathbb N}{\pi_{2k}}^{-1}\bigl(\{0\}\bigr).$$Since each ${\pi_{2k}}^{-1}\bigl(\{0\}\bigr)$ is closed, $Y$ is closed.
The space $Y^\perp$ is the space $Z$ of all sequence $(x_n)_{n\in\mathbb N}\in\ell^2$ such that $x_k=0$ whenever $k$ is odd. It is also clear that each element of $\ell^2$ can be written in one and only one way as the sum of an element of $Y$ and an element of $Z$. Therefore, $Z$ is indeed $Y^\perp$.
