Proof of equivalent characterizations of compact operators As an exercise I tried to prove the following theorem:

If $X,Y$ are Banach spaces and $u \in B(X,Y)$ is a bounded linear
  operator then the following are equivalent:
(1) $u$ is compact
(2) for every bounded set $S \subseteq X$ the image $u(S)$ is
  relatively compact
(3) if $x_n$ is a bounded sequence in $X$ then $u(x_n)$ admits a
  convergent subsequence in $Y$

Please could someone check my proof?
Proof:
Recall the definition of compact operator: $u$ is compact iff if the image of the unit ball is relatively compact. 
$(1) \iff (2)$: 
Since multiplication by $n$ is a linear homeomorphism, $u$ is compact iff $u(B(0,1))$ is relatively compact iff $u(B(0,n))$ is relatively compact. From this it is obvious that $(1) \iff (2)$.
$(2) \implies (3)$: Let $x_n$ be a bounded sequence. Then $S=\{x_n\}_{n \in \mathbb N}$ is a bounded set hence $u(S)$ is relatively compact hence $u(x_n)$ admits a subsequence that converges in $Y$. 
$(2) \Longleftarrow (3)$: Let $S$ be a bounded set. Let $x_n$ be any sequence in $\overline{u(S)}$. Then $x_n=u(s_n)$ for some sequence $s_n$ in $S$. Then $x_n$ admits a convergent subsequence. 
 A: The only problematic part of your proof is the implication $3)\implies 2)$. You should argue as follows
Assume $S$ is bounded. Consider arbitrary $(y_n)_{n\in\mathbb{N}}\subset\overline{u(S)}$, then $(y_n)_{n\in\mathbb{N}}$ is also bounded. For each $n\in\mathbb{N}$ we can find $x_n\in S$ such that 
$$
\Vert y_n-u(x_n)\Vert\leq 2^{-n}.\tag{*}
$$ Since $(y_n)_{n\in\mathbb{N}}$ is bounded, so does $(u(x_n))_{n\in\mathbb{N}}$. Since $3)$ holds, we have convergent subsequence $u(x_{n_k})_{k\in\mathbb{N}}$. From $(*)$ is follows that $(y_{n_k})_{k\in\mathbb{N}}$ converges to the same limit as $u(x_{n_k})_{k\in\mathbb{N}}$. Thus we constructed convergent subsequence of arbitrary sequence in $\overline{u(S)}$. So $u(S)$ is relatively compact.
A: The definition of a compact operator $u \in B(X,Y)$ is that $u$ maps bounded sets of $X$ into relatively compact sets of Y. Therefore, (1)  and (2) are equivalent by definition and no proof is required. This definition is equivalent to saying if $x_n$ is a bounded sequence in $X$, then $u(x_n)$ has a Cauchy subsequence (not necessarily convergent) in $Y$. To prove (3) $\Rightarrow$ (2), let $S$ be a bounded set in $X$. For any bounded sequence $x_n$ in $S$ there exists a Cauchy subsequence $u(x_{n_k}) \in u(S)$. Hence $u(S)$ is relatively compact.
Remember that relatively compactness means closure is compact. There may not exist a convergent subsequence (only definitely Cauchy).
