From a textbook:
Suppose $X$ and $Y$ are Banach spaces and $T : X \to Y$ is linear. Then $T$ is a compact operator if one of the following holds:
(a) $\{Tx_n\}$ contains a convergent subsequence in $Y$ whenever $\{x_n\}$ is a bounded sequene in $X$.
(b) The image of any bounded set $E$ in $X$ under $T$ has compact closure.
How can I show that (a) implies (b)?
Let $T:X\to Y$ be a continous linear operator and $E$ be a bounded set in $X$
Let's call $F = T(E)$,
Now the non obvious part is to show that $\overline{F} \subset T(\overline{E})$, this follow from the closed graph theorem
Let's assume this property, and take $(y_n)$ a sequence in $\overline{F}$
Now, for each $y_n$, let's take an $x_n \in \overline{E}$ such that $Tx_n = y_n$. $(x_n)$ is a bounded sequence, so $Tx_n = y_n$ has a converging subsequence, so $\overline{Y}$ is sequentially compact, hence compact.