Everywhere defined operators must be bounded? I have read in many places that as soon as you have an everywhere defined operator (on a Banach space), it must be automatically bounded, by the Closed Graph Theorem. However, I can't prove this using the Closed Graph Theorem (i.e., I can't prove it would be closed) and I can't find a reference for this. Is this true? Why?

You cannot prove that, as it is not true (with the axiom of choice). The statement, which is true from the closed graph theorem, is: 

If $T \colon X \to Y$ is a closed operator defined on a Banach space $X$ into a Banach space $Y$, than $T$ is bounded.


Addendum: Let $X$ be an infinite dimensional Banach space, $Y \ne 0$ be a Banach space. Then there is an unbounded $T \colon X \to Y$. Let (AC!) $B$ a basis of $X$ and $B' =\{b_n : n \in \mathbf N\}$ a countable subset, $y \in Y$ with $y \ne 0$. Define $T$ by linear extension of 
$$ T(b) = \begin{cases} n\|b_n\|y & b = b_n \\ 0 & b \in B \setminus B'\end{cases} $$
Then $T$ is linear $X \to Y$, and unbounded due to 
$$ \|T(b_n)\| = n\|b_n\|\|y\| $$
hence $\|T\| \ge n \|y\|$ for every $n$. 
