Let $X$ and $Y$ be Banach spaces and let $\Lambda : X \to Y$ be a bounded linear operator. Is it true that if $\Lambda$ is surjective and hence open, then $\Lambda(S)$ is closed for every closed set $S \subset X$ ?
$\textbf{Is the following a correct answer? if yes I wondered why we need the openness of $\Lambda$?}$
We have to show that any sequence $(x_n,y_n)$ in $\mathcal{G}(\Lambda (S)) = \{(x,y) : x\in S \ \ \& \ \ \ \Lambda x=y \in Y\}$ converges to a point $(x,y)$ in $\mathcal{G}(\Lambda (S))$.
Since $S$ is a closed subset of $X$, it is complete. So let's take any cauchy sequence $(x_n, \Lambda x_n)$ in $\mathcal{G}(\Lambda (S))$. Since $S\times Y$ is Banach, and $\Lambda$ is surjective we have that $(x_n, \Lambda x_n) \to (x,y)$ in $S\times Y$ for some $y$. It remains to show that $\Lambda x=y$.
We have that by completeness of $S$, that $x_n \to x$ in $S$, and by the continuity of $\Lambda$, $\Lambda x_n \to \Lambda x$, also we have that by definition of convergence $\Lambda x_n \to y$. So choose $N$ s.t. for all $n\geq N$, $\|\Lambda x_n - \Lambda x\|<\frac{\epsilon}{2}$ and $\|\Lambda x_n - y\|<\frac{\epsilon}{2}$. So
$$\|\Lambda x-y\|\le\|\Lambda x_n - \Lambda x\|+\|\Lambda x_n - y\|<\frac{\epsilon}{2}+\frac{\epsilon}{2} = \epsilon$$
so since $\mathcal{G}(\Lambda (S))$ is closed for any point in $S$, then $\Lambda (S)$ is also closed.
Let $X=\mathbb R^{2}$ and $Y=\mathbb R$. Let $\Lambda (x,y)=x$ and $S=\{(x,y)\in \mathbb R^{2}: xy=1\}$. Then the hypothesis is satisfied and $\Lambda (S)=\mathbb R \setminus \{0\}$ which is not closed.