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Is the set $\{x_n : n\in\mathbb{N}\}=\{\sin (nt): n\in\mathbb{N}\}\subset L_2[-\pi,\pi]$ closed, or further compact? I don't know how to prove it.

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Probably you should also say what topology you mean. The sequence $x_n$ converges to zero in the weak topology. – GEdgar Dec 10 '12 at 22:33

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It's not compact, because it is a discrete subspace of $L^2$, and there are no countable discrete compact spaces. You could also say that if it were compact, there'd be a convergent subsequence $x_{\phi(n)}$, but $x_n$ tends weakly to zero, and $0$ is not one of the $x_n$.

However, it is closed.

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The easiest way to see it is closed, is to remember the fact that the $\cos(nt)$ and $\sin(nt)$ together (and properly renormalised) form a Hilbert basis for $L^2$. From there on it is rather straightforward. – Olivier Bégassat Dec 10 '12 at 21:23
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Easier still: since it is discrete, it has no limit points, so vacuously it contains all its limit points, hence is closed. – Nate Eldredge Dec 10 '12 at 22:26
@NateEldredge I don't understand your comment, are you saying it is a closed subspace because it is discrete? That's not true, so what do you mean? – Olivier Bégassat Dec 11 '12 at 0:52
Sorry, I misused the word "discrete". What I meant was that it has no limit points, because as in Davide's answer, no pair of points is closer than distance 2. – Nate Eldredge Dec 11 '12 at 1:20
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As the elements are orthogonal, we have for $n\neq m$ that $$\lVert x_n-x_m\rVert_{L^2}^2=2,$$ proving that the set cannot be compact (it's not precompact, as the definition doesn't work for $\varepsilon=1/2$).

But it's a closed set, as it's the orthogonal of the even square-integrable functions.

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