Should continuity be a sufficient condition to proof Bounded linear map equivalent proposition?

Question

Since the problem is vague and my professor did not answer me. I just want to know whether the condition is sufficient to prove the following a) statement. Here is the original version.

$\text{Let ($\mathfrak X,\Vert\,\cdot\,\Vert_{\mathfrak X}$) and ($\mathfrak Y,\Vert\,\cdot\,\Vert_{\mathfrak Y}$) be normed linear spaces.}$ $\text{For $\textit{T}:$$\,\mathfrak X\,\rightarrow\,\mathfrak Y$ a linear map, we say that $\textit{T}$ is $\textbf{continuous}$ at a point $x\,\in V$ }$ $\text{if for each $\epsilon$>0 there exists $\delta$ >0 so that $\Vert x-y\Vert_{V}<\delta$ implies that $\Vert Tx-Ty\Vert_{W}<\epsilon.$}$
$\text {In other words, $T$ is continuous as a metric space map from $(\mathfrak X,d_{\mathfrak X})$ to $(\mathfrak Y,d_{\mathfrak Y})$.}$
(a) Suppose $T : \mathfrak X \rightarrow \mathfrak Y$ is bounded. Show that $$\Vert T\Vert =\inf\{K\in\mathbb R:\Vert Tx\Vert_{\mathfrak Y}\le K\Vert x\Vert_{\mathfrak X} \text{ for all }x\in \mathfrak X\}$$ My attempt is to move $\Vert x\Vert_{\mathfrak X}$ in the above set to the left hand side, and form a $K$'s inequility; however, I feel confused that for (a) there is no word says that $T$ is continuous, but in the introducing part, so my attempt will fail for those points whose norm tends to infinite. Since $\Vert Tx\Vert$ says nothing about those infinite norm point, hence it could be $\Vert Tx\Vert$$>>$$\Vert x\Vert$, even though $\Vert x\Vert $ is infinite, which means $\Vert Tx\Vert$ is higher order infinite to $\Vert x\Vert $. Hence, I cannot figure out other ways to prove this. Can someone tell me whether the continuity is necessary? If not, how to prove it? (If I know it is continuous, then I know it is pointwise bounded on its domain.) Thanks!
$\text{Here an operator is said to be bounded if its operator norm}$ $$\Vert T\Vert :=\sup\{\Vert Tx\Vert_{\mathfrak Y}:\Vert x\Vert_{\mathfrak X}\le1\}\lt\infty$$

Answer 1

For normed spaces, boundedness is equivalent to continuity. The reverse is tougher, but for a bounded operator $T$, $||Tx - Ty || = ||T(x-y)|| \leq ||T|| ||x-y||$. Which gives Lipschitz continuity with constant $||T||$.

edit: see https://en.wikipedia.org/wiki/Bounded_operator#Equivalence_of_boundedness_and_continuity for a proof in both directions.