Let $T:X\rightarrow Y$ be a bounded linear surjective operator where $X$ and $Y$ are Banach spaces and $y\in Y$. Then for $\{y_n\}$ converging to $y$, $\exists \{x_n\}$ converging to $x$ s.t. $T(x)=y$.
Edit: I meant there exists a convergent $\{x_n\}$ such that $T(x_n)=y_n$ and $x_n\rightarrow x$ as $n \rightarrow \infty$.
Best Answer
Its an application of the open mapping theorem. In particular a useful proposition is the following:
To prove this, from the open mapping theorem there is some $M'>0$ such that $M'B_Y\subseteq T(B_X)$ (i.e. $0$ belongs to the interior of $T(B_X)$), where $B_X=\{x\in X:\ ||x||\leq 1\}$ denotes the closed unit ball of $X$. Equivalently, $$\tag{1}B_Y\subseteq \frac{1}{M'}T(B_X)$$
Now pick $y\neq 0$ in $Y$. Then $y/||y||\in B_Y$, so from $(1)$ there is some $x'\in B_X$ such that $$\frac{1}{M'}Tx'=\frac{y}{||y||}$$
Then $T(\frac{||y||x'}{M'})=y$. Setting $x=\frac{||y||x'}{M'}$ we have $Tx=y$ and $||x||\leq \frac{1}{M'}||y||$. So if we pick $M=1/M'$ we have proved the above proposition. Now its easy to prove what you want since if $y_n\to y$. Then, fix some $x\in X$ with $Tx=y$.
Then use the proposition for the element $y_n-y$ and find $z_n\in X$ such that $Tz_n=y_n-y$ and $$\tag{2}||z_n||\leq M||y_n-y||$$
Then by $(2)$ it follows that $z_n\to 0$. Take $x_n=z_n+x$.