Q:

Closed graph theorem seems to state that a closed operator has to be bounded? 

By the closed graph theorem an operator $T$ is continuous  (equivalently bounded) if and only if it its graph is closed.  An operator with a closed graph is called a closed operator.  
So we have
$$
T \ \text{bounded} \Longleftrightarrow T \ \text{continuous} \Longleftrightarrow T \ \text{has closed graph} \Longleftrightarrow T \ \text{closed}.  \quad (*)
$$
But I often see closed operators mentioned in the context of unbounded operators.  That is unbounded operators can be closed.  But in $(*)$ above it seems that a closed operator is equivalent to a bounded operator? 

A:

Let $T : X\rightarrow Y$ be a linear operator from a Banach space $X$ to a Banach space $Y$. Then the closed graph theorem states that the following are equivalent

$T$ is continuous;
$T$ is bounded;
$T$ is closed. 

A linear operator $T : \mathcal{D}(T)\subsetneq X\rightarrow Y$ on a linear, dense domain $\mathcal{D}(T)$ does not satisfy the hypotheses of the closed graph theorem.  So $T$ can be closed without being continuous.  An example is the differentiation operator $T : C^1[0,1]\subset C[0,1]\rightarrow C[0,1]$. This operator is linear, densely-defined and closed, but is not bounded.