Compactness of $K(S)$, if $K$ is an infinite-dimensional compact operator in Hilbert space 
$H$ is infinite-dimensional Hilbert space. $K: H \rightarrow H$ is infinite-dimensional compact operator. Let $S$ be a unit sphere in H. The task is to proof that $K(S)$ couldn't be compact.

I know that:
Really we need to proof that $K[S]$ is not closed. If B is unit ball, $[B]$ - closure, then $K([B])$ is also closed(for compact operator in Hilbert space).
If $\mathrm{Ker} B$ is empty, i.e. $0 \notin K(S)$ then there is a sequence in $K(S)$ converging to $0$, so $K(S)$ isn't compact. But I don't understant how to proof noncompactness of $K(S)$ even if one of the basis vectors is mapped to $0$.
On the other side we can try to proof that if $K(S)$ is compact, then $\mathrm{Im}(K)=K(H)$ shuold have finite dimension. It is easy if $K(B)$ contains some ball(like a subspace in $K(H)$), but it's far from being true.
 A: In fact, $K(S)$ can be compact. Let $H = \ell^2(\mathbb{N})$, i.e.
$$H = \Biggl\{ f \colon \mathbb{N} \to \mathbb{C} \biggm\vert \sum_{k = 0}^{\infty} \lvert f(k)\rvert^2 < +\infty\Biggr\}$$
(using function notation to make indexing of sequences in $H$ clearer) and $K = C \circ D$, where
$$(Df)(k) = f(k+1)$$
and
$$(Cf)(k) = 2^{-k}f(k)\,.$$
Then $K(S)$ is closed, hence compact.
Let $g \in \overline{K(S)}$, and $(f_n)_{n \in \mathbb{N}}$ a sequence in $S$ with $K(f_n) \to g$. If $g = 0$, then we have $K(e_0) = g$ where $e_m(k) = \delta_{m,k}$, hence we can assume $g \neq 0$. For every fixed $k$ we have
$$f_n(k+1) \to 2^kg(k)$$
and therefore
$$\sum_{k = 0}^{m} 2^{2k}\lvert g(k)\rvert^2 = \lim_{n \to \infty} \sum_{k = 0}^{m} \lvert f_n(k+1)\rvert^2 \leqslant 1$$
for all $m$. It follows that $h \in H$, where
$$h(k) = 2^kg(k)\,.$$
Furthermore, we have $\lVert h\rVert \leqslant 1$, hence there is $h_1 \in S$  with $h = Dh_1$. By construction, $Ch = g$, thus
$$g = Kh_1 \in K(S)\,.$$
A: Actually, if $K \colon X\to Y$ is a compact operator and $X$ is reflexive, then   $K(S)$ is    compact.  Notice that since $K$ is compact, it must be completely continuous, i.e., if $x_n \to x$ weakly in $X$, then $||K(x_n)-K(x)||\to 0$.
(This is a corollary of  Banach-Steinhaus's Theorem.)
To see that  $K(S) $ is compact,  let $ (K(x_n))_n $ be a sequence in $K(S)$. Since $X$ is reflexive, $S$ is weakly compact and thus (by Eberlein-Smulian's Theorem), we can find a weakly convergent subsequence,  $x_{n_k} \to x \in X$ weakly. By the complete continuity of $K$, it follows that $K x_{n_k} \to Kx \in K(S)  $ (so, it has a (norm) convergent subsequence).
