Why is $\| T \frac{x}{\|x\|_E} \|_F \leq \|T \|$? Why is $\| T \frac{x}{\|x\|_E} \|_F \leq \|T \|$?
Where $T$ is a linear operator $T: E \rightarrow F$.
Intuitively $\frac{x}{\|x\|_E}=1$.
 A: If $||T||$ denotes the operator norm of $T$, then we know that
$$\left\lVert T\frac{x}{||x||_E}\right\lVert_F\leq||T||\cdot \left\lVert \frac{x}{||x||_E}\right\lVert_E=||T||\cdot\frac{1}{||x||_E}||x||_E=||T||,$$ by the operator norm inequality.
EDIT: It looks like the question is to prove the operator norm inequality (which makes more sense, since otherwise it's trivial). It seems like you might be confused on the last step, so I'll clarify it.
The operator norm inequality: $$||Tx||_F\leq ||T||\cdot ||x||_E.$$
Proof. If $x=0,$ then the result is obvious, so suppose not. Then,
$$\left\lVert T\left(\frac{x}{||x||_E}\right)\right\lVert_F\leq\sup_{||y||_E=1}||Ty||_F=||T||,$$  and so $$||Tx||_F\leq ||T|| ||x||_E.$$ This hinges on the fact that $$\left\lVert\frac{x}{||x||_E}\right\lVert_E=1,$$ and the operator norm involves a sup over all elements of norm $1$.
A: Recall how $||T||$ defined. Then, the inequality follows directly from the definition.
$$
||T||:=\sup_{||y||_E=1}||Ty||_F
$$
So, if you give a particular unit vector $y=x/||x||_E$, we have by the above definition 
$$
||T(x/||x||_E)||\leq\sup_{||y||_E=1}||Ty||_F=||T||
$$
