Let $U$ and $V$ be Banach spaces, let $A⊆U$ be a subspace, and let $f:A⊆U→V$ be an injective linear map with closed graph. Show that the inverse map $f^{-1} : B \to U$ is continuous if and only if $B = f(A)$ is closed in $V$.
For the forward implication, I basically did what was outlined here Closed Graph of a function and the continuous inverse implies the image is closed.
For the other direction, I need to show that if $B$ is closed in $V$, then whenever $b_i \to b \in B$, we have that $f^{-1}(b_i) \to f^{-1}(b)$. I don't really know how to begin.. as continuity requires $b_i \to b \in B$ to begin with, I don't see how to use the fact that $B$ is closed.
Let me denote the linear map by $T$ :)
Define the following norm on $A$ (graph norm of $T$): $$ \|x\|_A:= \|x\|_X + \|Tx\|_Y. $$ Then $(A,\|\cdot\|_A)$ is a Banach space. Moreover, $T: A\to T(A)$ is bijective, $(A,\|\cdot\|_A)$ and $(T(A),\|\cdot \|_Y)$ are Banach spaces, so $T^{-1}$ is continuous and bounded. Then there is a constant $c>0$ such that $$ \|y\|_Y \ge \|T^{-1}y\|_A = \|T^{-1}y\|_X + \|y\|_Y $$ for all $y\in T(A)$. This implies that $T^{-1}$ continuous from $T(A)$ to $(A,\|\cdot\|_X)$.