= Solution
The <open mapping theorem> states that every surjective bounded linear map $S:X\to Y$ between <Banach spaces> maps open sets to open sets. Equivalently, there is $c>0$ such that
$$
B_Y(0,c)\subseteq S(B_X(0,1)).
$$
The <closed graph theorem> states that a linear map $T:X\to Y$ between Banach spaces is continuous if and only if its <graph of a linear operator>[graph]
$$
\Gamma(T)=\{(x,Tx):x\in X\}
$$
is closed in $X\times Y$.
Assume the open mapping theorem and suppose $\Gamma(T)$ is closed. As a closed subspace of the Banach space $X\times Y$, the graph is itself Banach. The coordinate projection
$$
P:\Gamma(T)\longrightarrow X,\qquad (x,Tx)\longmapsto x
$$
is a bounded linear bijection. By the open mapping theorem, $P^{-1}$ is bounded. The other coordinate projection $Q:\Gamma(T)\to Y$ is bounded, and
$$
T=Q\circ P^{-1}.
$$
\b[Thus $T$ is bounded and hence continuous. This is the <closed graph theorem from the bounded inverse theorem> argument.]
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