I wanted to prove the following statement: Let $X,Y$ be Banach, $T:X\rightarrow Y$ be linear. Further for every sequence $(x_n)_{n \in \mathbb{N}}$ in X weakly convergent to $x\in X$, also $(Tx_n)_{n \in \mathbb{N}}$ converges weakly to $Tx$. Then T is continuous.
I know that you can prove the statement by contradiction assuming that T is not bounded. But I wondered if the following proof using the Closed Graph Theorem works as well?
Let $(x_n)_{n \in \mathbb{N}}$ in X such that $x_n \rightarrow x$ and $Tx_n \rightarrow y$. We want to show that $Tx = y$. Then T has closed graph and by the Closed Mapping Theorem T is continuous.
Now since $x_n \rightarrow x$ in norm it also converges weakly and hency by assumption also $(Tx_n)_{n \in \mathbb{N}}$ converges weakly to $Tx$. Additionally $Tx_n \rightarrow y$ implies that $(Tx_n)_{n \in \mathbb{N}}$ converges weakly to $y$. By the uniqueness of weak limits follows that $y=Tx$ and therefore the claim.