What can be said about the closedness and compactness of the subset $S = \{\sin (2^nx)\ |\ n \in \Bbb N \}$ in $L^2 ([-\pi, \pi] )\ $? 
Let $L^2 \left ([-\pi,\pi ] \right )$ be the metric space of Lebesgue square integrable functions on $[-\pi, \pi]$ with a metric $d$ given by $$d(f,g) = \left [\int_{-\pi}^{\pi} \left (f(x) - g(x) \right )^2\ dx \right ]^{\frac {1} {2}},\ \text {for}\ f, g \in L^2 \left ([-\pi, \pi] \right ).$$ Consider the subset $$S = \left \{\sin \left (2^nx \right )\ |\ n \in \Bbb N \right \}\ \text {of}\ L^2 \left ([-\pi, \pi ] \right ).$$ Which of the following statements are true?


$(1)$ $S$ is bounded


$(2)$ $S$ is closed


$(3)$ $S$ is compact


$(4)$ $S$ is non-compact

It is easy to show that $S$ is a compact subset of $\left (L^2 \left ([-\pi, \pi] \right ), d \right ).$ But I can't able to prove or disprove whether it is closed or not. Let $\{x_n \}_{n \geq 1}$ be a convergent sequence in $S.$ Then there exists a strictly increasing sequence of natural numbers $\{k_n\}_{n \geq 1}$ such that $x_n = \sin \left (2^{k_n} x \right ),\ \text {for all}\ n \geq 1.$ Let it converge to some $f \in L^2 \left ([-\pi, \pi] \right ).$ So we have $$\int_{-\pi}^{\pi} \left (\sin \left (2^{k_n} x \right ) - f(x) \right )^2\ dx \to 0\ \text {as}\ n \to \infty.$$ From here how do I conclude that $f \in L^2 \left ([-\pi, \pi ] \right )\ $? Also how to conclude about the compactness or non-compactness of $S\ $? Any help in this regard will be appreciated.
Thanks for your time.
 A: As you mention, the set $S$ itself has no limit point : for every $m \neq n$ , we have $d(\sin(2^n x), \sin(2^m x)) = \sqrt{2 \pi}$. Therefore, the sequence $\sin(2^n x), n=1,2,3,...$ will not have a convergent subsequence, as no subsequence is even Cauchy.
In this regard, certainly $S$ is not compact. (It is not sequentially compact, to be precise, but these mean the same in our space $L^2[-\pi,\pi]$ which is second countable).
However, $S$ is closed. This is because if you take a Cauchy sequence in $S$, then by what we have said it has to be an eventually constant sequence, so of course it converges with the limit in $S$.
Furthermore, $S$ being bounded is obvious, since $|\sin (2^nx)| \leq 1$ for all $n$ and $x$ so the norm of any element in $S$ is just bounded by $\sqrt {2 \pi}$.

Finally, $S$ is a closed, bounded and non-compact subset of $L^2[-\pi,\pi]$. Contrast with the Heine-Borel theorem.

EDIT
We have for any $k \neq l$ positive integers that $$
(\sin kx - \sin lx)^2 = \sin^2 kx + \sin^2 lx - 2 \sin kx \sin lx   \\
= \sin^2 kx + \sin^2 lx - 2(\cos(k-l)x - \cos(k+l)x) \\  
= \frac{1 - \cos 2kx}{2} + \frac{1 - \cos 2 lx}{2} - 2(\cos(k-l)x - \cos(k+l)x)
$$
we have for any $t \neq 0$ integer that $\int_{- \pi}^{\pi} \cos(tx) = \left[\frac{\sin(tx)}{t}\right]_{-\pi}^{\pi} = 0 - 0 = 0$. Therefore, provided that $k \neq l$ above, when we integrate $(\sin kx - \sin lx)^2$ from $- \pi$ to $\pi$ only the constants remain as all the cosines integrate to $0$. The constant is just $\frac 12 + \frac 12 = 1$, so the integral is just $\int_{-\pi}^{\pi} 1dx = 2 \pi$.
Set $k = 2^m$ and $l= 2^n $ for $m \neq n$ to see our case. Also we must take the square root since $d$ has a square root, so the final answer is $\sqrt{2 \pi}$.
