If for every $x_n$ such that $x_n \rightarrow x$, there exists a $x_{n_k}$ such that $Tx_{n_k} = Tx$, is $T$ continuous? Let $X$ and $Y$ be Banach spaces and $T$ be the (possibly nonlinear) map $T\colon X \rightarrow Y$. $T$ is continuous if for every $x_n \in X$ such that $x_n \rightarrow x$, then $Tx_n \rightarrow Tx$. Is $T$ also continuous if instead, for every $x_n \in X$ such that $x_n \rightarrow x$, then there exists a subsequence $x_{n_k}$ such that $Tx_{n_k} = Tx$?

I misunderstood the question first. Indeed the answer is yes. If $Tx_n$ does not go to $Tx$ one can choose subsequence $x_{n_k}$ that is bounded away from $Tx,$ e.g. $\|Tx_{n_k}-Tx\|\ge\varepsilon.$ This leads to a contradiction.    
