Why is injectiveness needed to show that a bounded linear operator is a closed map?

Question

Let $X$ and $Y$ be Banach. I'm trying to show that a bounded linear map $T:X\rightarrow Y$ is a closed map, that is $A\subset X$ is closed $\implies T(A)\subset Y$ is closed, if $T$ is injective and $T(X)$ is closed. My question is: Is the injectiveness of $T$ necessary?$$ $$ If we consider a sequence $\left(x_n\right)_{n\in\mathbb{N}}\subset A$ such that $\lim\limits_{n \rightarrow \infty}{x_n}=x$ for some $x\in X$, then $x\in A$, because $A$ is closed. So by continuity of $T$ we get $\lim\limits_{n \rightarrow \infty}{Tx_n}=Tx\in T(A)$. I don't see where it is necessary to assume that $T$ is injective.