Clarification on proof that for a linear map T, being continuous and being Lipschitz are equivalent statements I know that in general for any function $f:V \longrightarrow W $ from a normed vector space $V$ to a second normed vector space $W$, being a Lipschitz function implies being continuous, is the second implication on my question the one I'm having troubles to understand.
 A: Hint: Consider an open ball around $0$ within $f^{-1}(B_1(0))$ to obtain that $f$ is bounded and use linearity.
A: If $\dim V$ and $\dim W$ are finite, here it is a proof for the sake of curiosity.
Based on such assumption, let us consider a basis $\mathcal{B} = \{b_{1},b_{2},\ldots,b_{n}\}$ for $V$ and a basis $\mathcal{B}' = \{b'_{1},b'_{2},\ldots,b'_{m}\}$ for $W$, where $n = \dim V$ and $m = \dim W$.
Then one has the following matricial representation for $T$:
\begin{align*}
[T]_{\mathcal{B}}^{\mathcal{B}'} = [[T(b_{1})]_{\mathcal{B}'}\, [T(b_{2})]_{\mathcal{B}'} \ldots\,[T(b_{n})]_{\mathcal{B}'}]
\end{align*}
Consequently, if $v = a_{1}b_{1} + a_{2}b_{2} + \ldots + a_{n}b_{n}$, one has that
\begin{align*}
T(v) = [T]_{\mathcal{B}}^{\mathcal{B}'}[v]_{\mathcal{B}} = a_{1}[T(b_{1})]_{\mathcal{B'}} + a_{2}[T(b_{2})]_{\mathcal{B'}} + \ldots + a_{n}[T(b_{n})]_{\mathcal{B'}} 
\end{align*}
Finally, according to the triangle inequality as well as the Cauchy-Schwarz inequality, we have that
\begin{align*}
\|T(v)\| & = \|a_{1}[T(b_{1})]_{\mathcal{B'}} + a_{2}[T(b_{2})]_{\mathcal{B'}} + \ldots + a_{n}[T(b_{n})]_{\mathcal{B'}} \|\\\\
& \leq \|a_{1}[T(b_{1})]_{\mathcal{B'}}\| + \|a_{2}[T(b_{2})]_{\mathcal{B'}}\| + \ldots + \|a_{n}[T(b_{n})]_{\mathcal{B'}}\|\\\\
& = |a_{1}|\|[T(b_{1})]_{\mathcal{B'}}\| + |a_{2}|\|[T(b_{2})]_{\mathcal{B'}}\| + \ldots + |a_{n}|\|[T(b_{n})]_{\mathcal{B'}}\|\\\\
& = \langle(|a_{1}|,|a_{2}|,\ldots,|a_{n}|),(\|[T(b_{1})]_{\mathcal{B'}}\|,\|[T(b_{2})]_{\mathcal{B'}}\|,\ldots,\|[T(b_{n})]_{\mathcal{B'}}\|)\rangle \leq K\|v\|
\end{align*}
Finally, due to the linearity of $T$, the desired result holds:
\begin{align*}
\|T(v) - T(w)\| = \|T(v - w)\| \leq K\|v - w\|
\end{align*}
Hopefully this helps!
