Prob. 6, Sec. 3.3, in Kreyszig's Functional Analysis Book: What is $Y^\perp$ if $Y = \ldots \subset \ell^2$? 
Let $Y$ be the subset of $\ell^2$ given by 
  $$ Y \colon= \left\{ \ \left( \xi_n \right)_{n \in \mathbb{N}} \in \ell^2 \ \colon \ \xi_{2n} = 0 \ \mbox{ for all } \ n \in \mathbb{N} \ \right\}. $$
  Then $Y$ is a closed subspace of the Hilbert space  $\ell^2$, and so $Y$ is also complete. What is $Y^\perp$? 

Here is the relevant definition: 

Let $X$ be an inner product space, and let $M$ be a non-empty subset of $X$. Then the orthogonal complement $M^\perp$ of $M$ is defined as follows: 
  $$ M \colon= \left\{ \ x \in X \ \colon \  \langle x, y \rangle = 0 \ \mbox{ for all } \ y \in M \ \right\}. $$

My effort: 

By definition, 
  \begin{align}
& Y^\perp \\
&= \left\{ \ x \in \ell^2 \ \colon \ \langle x, y \rangle = 0 \ \forall y \in Y \ \right\} \\
&= \left\{ \ \left( \xi_n \right)_{n \in \mathbb{N} } \in \ell^2 \ \colon \ \left\langle \left( \xi_n \right)_{n \in \mathbb{N} }, \left( \eta_n \right)_{n \in \mathbb{N} } \right\rangle = 0 \ \forall \left( \eta_n \right)_{n \in \mathbb{N} } \in Y \ \right\} \\
&= \left\{ \ \left( \xi_n \right)_{n \in \mathbb{N} } \in \ell^2 \ \colon \ \sum_{n = 1}^\infty  \xi_n  \overline{  \eta_n} = 0 \ \forall \left( \eta_n \right)_{n \in \mathbb{N} } \in Y \ \right\} \\ 
&= \left\{ \ \left( \xi_n \right)_{n \in \mathbb{N} } \in \ell^2 \ \colon \ \sum_{n = 1}^\infty  \xi_n  \overline{  \eta_n} = 0 \ \forall \left( \eta_n \right)_{n \in \mathbb{N} } \in \ell^2 \ \mbox{ such that } \ \eta_{2n} = 0 \ \forall n \in \mathbb{N}  \ \right\} \\
&= \left\{ \ \left( \xi_n \right)_{n \in \mathbb{N} } \in \ell^2 \ \colon \ \sum_{n = 1}^\infty  \xi_{2n-1}  \overline{  \eta_{2n-1} } = 0 \ \forall \left( \eta_n \right)_{n \in \mathbb{N} } \in \ell^2 \ \mbox{ such that } \ \eta_{2n} = 0 \ \forall n \in \mathbb{N}  \ \right\}. 
\end{align}
  My intuition is that 
  $$Y^\perp = \left\{ \ \left( \xi_n \right)_{n \in \mathbb{N} } \in \ell^2 \ \colon \ \xi_{2n-1} = 0 \ \forall n \in \mathbb{N}  \ \right\}.$$ 

Is it so? If yes, then how to prove this rigorously? 
Another approach: 

As $\ell^2$ is a Hilbert space and as $Y$ is a closed subspace of $\ell^2$, so $$ \ell^2 = Y \oplus Y^\perp, $$
  by Theorem 3.3-4 in the book Introductory Functional Analysis With Applications by Erwine Kreyszig. 
This means that every $x \in \ell^2$ has a unique representation 
  $$ x = y+z, \ \mbox{ where } \ y \in Y \ \mbox{ and } \ z \in Y^\perp. $$
Now  if $x \colon= \left( \xi_n \right)_{n \in \mathbb{N}} \in \ell^2$, then, by definition, the series $\sum \left\lvert \xi_n \right\rvert^2$ of non-negative real numbers converges in $\mathbb{R}$. 
Therefore, the sequence 
  $$ \left( \sum_{j=1}^n \left\lvert \xi_j \right\rvert^2 \right)_{n \in \mathbb{N}} $$ of the partial sums of the series $\sum \left\lvert \xi_n \right\rvert^2$, which is also monotonically increasing, is convergent. Hence this sequence is bounded (from above). That is, for each  $n \in \mathbb{N}$, we have 
  $$ \sum_{j=1}^n \left\lvert \xi_j \right\rvert^2 \leq \sum_{j=1}^{n+1} \left\lvert \xi_j \right\rvert^2, $$
  and 
  $\lim_{n \to \infty} \sum_{j=1}^n \left\lvert \xi_j \right\rvert^2$ exists in $\mathbb{R}$, and
$$ \lim_{n \to \infty} \sum_{j=1}^n \left\lvert \xi_j \right\rvert^2 = \sup \left\{ \ \sum_{j=1}^n \left\lvert \xi_j \right\rvert^2 \ \colon \ n \in \mathbb{N} \ \right\} < +\infty. $$
Now let $y \colon= \left( \eta_n \right)_{n \in \mathbb{N} }$ be ths sequence of (real or complex) numbers defined as follows: For each $n \in \mathbb{N}$, let 
  $$ \eta_{2n-1} \colon= \xi_{2n-1} \ \mbox{ and } \eta_{2n} \colon= 0. $$
  Then the sequence 
  $$ \left( \sum_{j=1}^n \left\lvert \eta_j \right\rvert^2 \right)_{n \in \mathbb{N} } $$ 
  of the partial sums of the series $\sum \left\lvert \eta_j \right\rvert^2$ is given by 
  $$ \sum_{j=1}^{2n-1} \left\lvert \eta_j \right\rvert^2 = \sum_{j=1}^n \left\lvert \xi_{2j-1} \right\rvert^2, $$
  and 
  $$ \sum_{j=1}^{2n} \left\lvert \eta_j \right\rvert^2 = \sum_{j=1}^n \left\lvert \xi_{2j-1} \right\rvert^2 $$
  for all $n \in \mathbb{N}$. And, this sequence, being a sequence of sums of non-negative real numbers, is also monotonically increasing and  bounded above by the sum of the convergent series $ \sum \left\lvert \xi_n \right\rvert^2$. So the sequence $$ \left( \sum_{j=1}^n \left\lvert \eta_j \right\rvert^2 \right)_{n \in \mathbb{N} } $$ is convergent, which means that the series 
  $ \sum  \left\lvert \eta_j \right\rvert^2$ converges. Therefore, 
   $y \colon= \left( \eta_n \right)_{n \in \mathbb{N} }$ is in $\ell^2$. 
  So from the definition of $Y$ we can conclude that $y \in Y$. 
Now let $z \colon= \left( \zeta_n \right)_{n \in \mathbb{N}}$ be the sequence of (real or complex) numbers defined as follows: For each $n \in \mathbb{N}$, let 
  $$ \zeta_{2n-1} \colon= 0 \ \mbox{ and } \ \zeta_{2n} \colon= \xi_{2n}. $$
  Then $x = y+z$, and so $z = x-y \in \ell^2$, because $\ell^2$ is closed under addition and scalar multiplication. 
Another way to show that $z \colon= \left( \zeta_n \right)_{n \in \mathbb{N}}$ is in $\ell^2$ is to show that the series $\sum \left\lvert \zeta_n \right\rvert^2$ of real numbers converges in $\mathbb{R}$. 
Now the sequence 
  $$ \left( \sum_{j=1}^n  \left\lvert \zeta_j \right\rvert^2 \right)_{n \in \mathbb{N} }$$ 
  of the partial sums of the series $\sum \left\lvert \zeta_n \right\rvert^2$ is  given by 
  $$ \sum_{j=1}^{2n-1}  \left\lvert \zeta_j \right\rvert^2 = \sum_{j=1}^{2n-2}  \left\lvert \xi_{2j} \right\rvert^2,   $$ 
  and 
  $$ \sum_{j=1}^{2n}  \left\lvert \zeta_j \right\rvert^2 = \sum_{j=1}^{2n}  \left\lvert \xi_{2j} \right\rvert^2   $$ 
  for all $n \in \mathbb{N}$. Thus this sequence, being a sequence of sums of non-negative real numbers, is monotonically increasing and bounded above by the sum of the series $\sum \left\lvert \xi_n \right\rvert^2$, and  so this sequence is convergent, which implies that $z \in \ell^2$. 
Thus 
  $$ y = \left( \eta_1, \eta_2, \eta_3, \ldots \right) = \left( \xi_1, 0, \xi_3, 0, \xi_5, 0, \ldots \right), $$ 
  and 
  $$ z = \left( \zeta_1, \zeta_2, \zeta_3, \ldots \right) = \left( 0, \xi_2, 0, \xi_4, 0, \xi_6,  \ldots \right),  $$ 
  and so 
  $$ \langle y, z \rangle = \sum_{n=1}^\infty \eta_n \overline{\zeta_n} = 0, $$
  and so $\langle y, z \rangle = 0$ also. In fact, we can also show that 
  $\langle z, v \rangle = 0$ for all $v \in Y$. Thus $z \in Y^\perp$. 
But so far I have only been able to show that the subspace 
  $$ Z = \left\{  \ \left( \xi_n \right)_{n \in \mathbb{N} }  \in \ell^2 \ \colon \ \xi_{2n-1} = 0 \ \forall \ n \in \mathbb{N} \ \right\}$$ is contained in $Y^\perp$. 

Is what I have done so far correct, logically sound, and rigorous enough? 
If so, then how to complete the above two arguments? 
If not, then where do I lack? 
P.S.:

In order to show that $Y$ is indeed closed in $\ell^2$, let us suppose that $x \colon= \left( \xi_n \right)_{n \in \mathbb{N} }$ is an element in the closure of $Y$ in $\ell^2$. Then, by Theorem 1.4-6 (a) in Kreyszig, there exists a sequence $\left(y_m \right)_{m \in \mathbb{N} }$, where $y_m \colon= \left( \eta_{mn} \right)_{n \in \mathbb{N} }$ for each $m \in \mathbb{N}$, converging to $x$ in $\ell^2$. 
For each $m \in \mathbb{N}$, as $y_m = \left( \eta_{mn} \right)_{n \in \mathbb{N} } \in Y$, so we must have $\eta_{m, 2n} = 0$ for all $n \in \mathbb{N}$, that is, we must have $$ \eta_{m2} = \eta_{m4} = \eta_{m6} = \cdots = 0. $$
Now let us choose an arbitrary natural number $N$. We show that the sequence $\left( \eta_{mN} \right)_{m \in \mathbb{N} }$ converges (in $\mathbb{R}$ or $\mathbb{C}$ as the case may be) to the number $\xi_N$.
As the sequence $\left(y_m \right)_{m \in \mathbb{N} }$ converges in $\ell^2$ to the point $x$, so given any real number  $\varepsilon > 0$, there exists a natural number $M$ such that 
  $$ d_{\ell^2} \left( y_m, x \right) < \varepsilon $$
  for all $m \in \mathbb{N}$ such that $m > M$. 
  But we have $x = \left( \xi_n \right)_{n \in \mathbb{N} }$, and, for each $m \in \mathbb{N}$, we have $y_m = \left( \eta_{m n} \right)_{n \in \mathbb{N} }$. So we can conclude that 
  $$ \sqrt{ \sum_{n = 1}^\infty \left\lvert \eta_{m n} - \xi_n \right\rvert^2 } < \varepsilon $$
  for all $m \in \mathbb{N}$ such that $m > M$.
And, so we obtain
  $$ \left\lvert \eta_{mN} - \xi_N \right\rvert = \sqrt{ \left\lvert \eta_{mN} - \xi_N \right\rvert  } \leq \sqrt{ \sum_{n = 1}^\infty \left\lvert \eta_{mn} - \xi_n \right\rvert^2 } < \varepsilon $$
  for all $m \in \mathbb{N}$ such that $m > M$. Thus it follows that the sequence $\left( \eta_{m N} \right)_{m \in \mathbb{N} }$ converges to $\xi_N$. 
Now since for each $N \in \mathbb{N}$, we have 
  $$ \xi_N = \lim_{m \to \infty} \eta_{m N}, $$
  and since for each $N \in \mathbb{N}$, we have $\eta_{m, 2N} = 0$, 
  therefore we must have
  $$ \xi_{2N} = 0$$
  for all $N \in \mathbb{N}$, which shows that $x = \left( \xi_{n} \right)_{n \in \mathbb{N} }$ is in $Y$. 
But this $x$ was an arbitrary element of the closure of $Y$ in $\ell^2$. Thus it follows that $\mbox{Cl}(Y) \subset Y \subset \mbox{Cl}(Y)$. Hence $Y$ is closed.

 A: It is correct, but also making it more complicated than it has to be. You want to show that
$$ Z = \left\{  \ \left( \xi_n \right)_{n \in \mathbb{N} }  \in \ell^2 \ \colon \ \xi_{2n-1} = 0 \ \forall \ n \in \mathbb{N} \ \right\}$$
is $Y^\perp$. As you wrote, it is clear that $Z\subset Y^\perp$. On the other hand $Z^c\subset (Y^\perp)^c$: take any $z\notin Z$ and take $n$ such that $z_{2n-1}\neq0$ (which exists by definition). Consider $e(2n-1)$, the vector with all $0$-s except at the $2n-1$-th coordinate, where it is $1$. Of course $e(2n-1)\in Y$ and $\langle z,e(2n-1)\rangle=z_{2n-1}\neq0$, so $z\notin (Y^\perp)^c$.
A: You have the correct intuition.  We should indeed find that 
$$
Y^\perp = \left\{ \ \left( \xi_n \right)_{n \in \mathbb{N} } \in \ell^2 \ \colon \ \xi_{2n-1} = 0 \ \forall n \in \mathbb{N}  \ \right\}.
$$
With the first approach, you've already showed that
$$
Y^\perp = \left\{ \ x \in \ell^2 \ \colon \ \langle x, y \rangle = 0 \ \forall y \in Y \ \right\}
\\= \left\{ \ \left( \xi_n \right)_{n \in \mathbb{N} } \in \ell^2 \ \colon \ \sum_{n = 1}^\infty  \xi_{2n-1}  \overline{  \eta_{2n-1} } = 0 \ \forall \left( \eta_n \right)_{n \in \mathbb{N} } \in \ell^2 \right\}
$$
(note that because we reindexed the sum, there is no need to say $\eta_{2n} = 0$ for all $n$).
Now, let $S = \left\{ \ \left( \xi_n \right)_{n \in \mathbb{N} } \in \ell^2 \ \colon \ \xi_{2n-1} = 0 \ \forall n \in \mathbb{N}  \ \right\}$.  It's easy to show that $S \subset Y^\perp$: if $\xi_{2n-1} = 0$ for all $n$, then $\sum_{n = 1}^\infty  \xi_{2n-1}  \overline{  \eta_{2n-1} } = 0$ for any $(\eta_n)$.  
To show that $Y^\perp \subset S$, take any $(\xi_n) \in Y^\perp$.  define $\eta_n$
$$
\eta_n = \begin{cases}
\xi_n & n \text{ is even}\\
0 & n \text{ is odd}
\end{cases}
$$
We then have 
$$
\langle (\xi_n),(\eta_n) \rangle = 0 \implies\\
\sum_{n=1}^\infty |\xi_n|^2 = 0 \implies (\xi_n) \in S
$$
as desired.

You might find that it's easier to do the problem, however, by first showing that the set
$$
\{e_{2n} : n \in \Bbb N\}
$$
(as in section 2.3) forms a Schauder-basis for the space $Y$.
