Proving that $\Vert T \Vert =\sup\limits_{\Vert x \Vert\leq 1}\Vert T x \Vert= \sup\limits_{\Vert x \Vert = 1}\Vert T x \Vert \cdots$

Let $$B(X,Y)$$ be the family of all bounded maps from $$X$$ to $$Y.$$ Then, we have for arbitrary $$T\in B(X,Y),$$ \begin{align} \Vert T \Vert =\sup\limits_{\Vert x \Vert\leq 1}\Vert T x \Vert= \sup\limits_{\Vert x \Vert = 1}\Vert T x \Vert=\sup\limits_{x\neq 0} \frac{\Vert T x \Vert}{\Vert x \Vert}.\end{align}

Proof

Since $$T$$ is bounded and linear, $$\exists K\geq 0$$ such that for all $$x\in X,\;\Vert T x \Vert \leq K \Vert x \Vert.$$ If $$\Vert x \Vert\leq 1$$, then $$\Vert T x \Vert \leq K \Vert x \Vert\leq K.$$

Thus, \begin{align}\tag{1}\label{1} \sup\limits_{\Vert x \Vert\leq 1}\Vert T x \Vert\leq \inf\{ K\geq 0:\Vert T x \Vert \leq K \Vert x \Vert,\forall\;x\in K \}=\Vert T \Vert\end{align} By definition of $$\inf,$$ for every $$\epsilon> 0,\exists \;x_{\epsilon}\in X,x_{\epsilon}\neq 0$$ such that \begin{align} \Vert T x_{\epsilon} \Vert>\left(\Vert T \Vert -\epsilon\right)\Vert x_{\epsilon} \Vert.\end{align} Let $$u_{\epsilon}=\frac{x_{\epsilon}}{\Vert x_{\epsilon} \Vert},$$ then $$u_{\epsilon}=1$$ and $$\Vert T u_{\epsilon} \Vert>\Vert T \Vert -\epsilon.$$ We obtain from $$\eqref{1}$$

\begin{align} \tag{2}\label{2}\Vert T \Vert \geq\sup\limits_{\Vert x \Vert\leq 1}\Vert T x \Vert\stackrel{\text{how?}}{\geq} \sup\limits_{\Vert x \Vert = 1}\Vert T x \Vert\stackrel{\text{how?}}{\geq} \sup\limits_{\Vert x_{\epsilon} \Vert\neq 0} \Vert T\left(\frac{x_{\epsilon} }{\big \Vert x_{\epsilon} \Vert}\right)\big\Vert\stackrel{\text{how?}}{\geq}\Vert T \Vert -\epsilon.\end{align} Since $$\epsilon>0$$ was arbitrary, then \begin{align} \Vert T \Vert =\sup\limits_{\Vert x \Vert\leq 1}\Vert T x \Vert= \sup\limits_{\Vert x \Vert = 1}\Vert T x \Vert=\sup\limits_{x\neq 0} \frac{\Vert T x \Vert}{\Vert x \Vert}.\end{align} Can you please explain the how's in $$\eqref{2} ?$$

The first one is because you're taking the sup over a smaller set. When $$A \subset B$$, and $$f$$ is a real valued map whose domain contains $$A$$ and $$B$$, we have $$\sup_{x \in A} f(x) \le \sup_{x \in B} f(x).$$ Here $$B = \{x \, : \, \|x\| \le 1\}$$ and $$A = \{ x \, : \, \|x \| = 1\}$$.
The second is again just restricting to a smaller set. For each $$\epsilon$$, the vector $$x_\epsilon/\|x_\epsilon\|$$ has norm $$1$$, so if we only consider the sup over those vectors, we are again considering a smaller set, and thus get the inequality for the same reason as above.
The third is by the definition of $$x_\epsilon$$. We've chosen $$x_\epsilon$$ such that $$\left \| T\left(\frac{x_\epsilon}{\| x_\epsilon\|} \right)\right \| \ge \|T \| - \epsilon.$$
1. Since $$\bigl\{\lVert Tx\rVert\,|\,\lVert x\leqslant1\bigr\}\supset\bigl\{\lVert Tx\rVert\,|\,\lVert x=1\bigr\}$$, $$\sup\bigl\{\lVert Tx\rVert\,|\,\lVert x\leqslant1\bigr\}\leqslant\sup\bigl\{\lVert Tx\rVert\,|\,\lVert x=1\bigr\}$$
2. Because $$\left\lVert\dfrac{x_\varepsilon}{\lVert x_\varepsilon\lVert}\right\rVert=1$$.
3. It was proved before that $$\left\lVert T\left(\dfrac{x_\varepsilon}{\lVert x_\varepsilon\lVert}\right)\right\rVert\geqslant\lVert T\rVert-\varepsilon$$.