A question about Banach spaces and closed graph theory 
Let $X, Y$ be Banach spaces on the same field.
Let $T \in B(X,Y)$ a bijection (surjective and injective). Where $B(X,Y)$ denotes the collection of the linear, bounded operators.
Prove that: $\inf_{\|x\|=1} \|Tx\| > 0$.

How to use the closed graph theorem to show this?
 A: Since $T:X\to Y$ is a bijection and $X,Y$ Banach spaces it follows that $T^{-1}:Y\to X$ it is also continuous. Indeed, we can show that $T^{-1}$ is continuous by showing that $T^{-1}$ has closed graph. So, suppose $T^{-1}y_n\to x$ and $y_n\to y$. We wish to show that $T^{-1}y=x$. By the continuity of $T$, $y_n\to Tx$. But, $y_n\to y$ also. Hence, $y=Tx$ and it follows that $T^{-1}y=x$. Hence, $T^{-1}$ is continuous. It follows that
$$||T^{-1}y||\leq ||T^{-1}||\cdot ||y||$$
Plugging $y=Tx$ we obtain
$$||x||\leq ||T^{-1}||\cdot ||Tx||$$
for every $x$. Now, $T^{-1}\neq 0$ implies that $||T^{-1}||>0$ (otherwise $X=Y=\{0\})$ Hence,
$$||Tx||\geq \frac{||x||}{||T^{-1}||}$$
for every $x$. In particular,
$$\inf_{||x||=1}||Tx||\geq \frac{1}{||T^{-1}||}>0$$
A: The data given does not immediately imply that $T$ is a homeomorphism: the inverse of a continuous bijection may be discontinuous, so what you write is not quite right. In any case, what you write does not really lead you in the right direction (it would only give you the opposite inequality, which is more or less given).
Here's an outline of a correct proof:

*

*$T$ is continuous (because it is bounded).

*The graph of $T$ is closed, as you say (because $Y$ is a Hausdorff space).

*This implies that the graph of $T^{-1}$ is also closed.

*By the closed graph theorem, you can conclude that $T^{-1}$ is continuous.

*This implies that $T^{-1}$ is bounded, which can be used to obtain what you need.

