functional analysis - $T$ compact operator implies that every weakly convergent sequence $x_n$ implies strong convergence of $Tx_n$

I'm working on a problem regarding compact operators and weakly convergent sequences. We know that an operator $T$ on a Hilbert space $H$ is compact iff for every bounded sequence $(x_n)_n \subset H$ its image $(Tx_n)_n \subset H$ admits a convergent subsequence (one can use this as a definition of a compact operator. I wanna show the following:

Let $T$ be a compact operator on $H$. Then for every sequence $(x_n)_n \subset H$ which converges weakly to $0$, the sequence $(Tx_n)_n)$ converges to $0$ in norm.

Can someone help me?