A bounded operator satisfies $\|Tx\|\leq\|T\| \|x\|$ I want to prove that if $T$ is a bounded linear operator, from $X$ to $Y$, then for each $x\in X $ we must have  $\|Tx\|\leq\|T\| \|x\|$.
Let's take some nonzero $x\in X$ . Then $\frac{\|Tx\|}{\|x\|} \leq \sup_x \frac{\|Tx\|}{\|x\|} $. But this supremum is the definition of $\|T\|$ so we must have that $\|Tx\| \leq \|T\| \|x\|$. I'm not sure if I can take the sup as I did, and with $x$ changing in the denominator, and in the image, that it will be necessarily bigger or equal to the right hand side.

If you take the definition of $\|T\|$ to be $\|T\| = \sup\left\{\dfrac{\|Tx\|}{\|x\|} : x \in X \right\}$, then it is a triviality that $\|Tx\| \le \|T\|\|x\|$ for every $x\in X$.
But sometimes the definition is taken to be $\|T\| = \sup\left\{\dfrac{\|Tx\|}{\|x\|} : \|x\|\le 1 \right\}$ or $\|T\| = \sup\left\{\dfrac{\|Tx\|}{\|x\|} : \|x\|\ = 1 \right\}$.  In those cases there's just a bit more work.  You can write
$$
\|Tx\| = \left\| \|x\| T\left( \frac x {\|x\|} \right) \right\| = \|x\| \left\| T \frac x {\|x\|} \right\| \le \|x\| \|T\| \left\| \frac x {\|x\|}\right\| = \|x\|\|T\| \frac{\|x\|}{\|x\|} = \|x\|\|T\|.
$$
