I came across this question in my self study of a little fuctional analysis and I'd help in solving it. I have also attempted it in the hope that I'll be corrected where I made mistakes. Thanks.
Let $X$ and $Y$ be normed linear spaces and $T:X\to Y$ be linear. I want to show the following:
(a) $T$ is continuous if and only if it is continuous at a single point $x_0$ in $X$.
(b) $T$ is Lipschitz if and only if it is continuous.
(c) Neither (i) nor (ii) hold in the absence of the linearity assumption on $T$.
Attempts.
(a)
$(\Rightarrow)$ Obvious, since if $T$ is continuous everywhere in $X$, then it is certainly continuous at a point $x_0\in X$.
$(\Leftarrow)$ Suppose that $T$ is continuous at $x_0 \in X$. Let $\varepsilon \gt 0$ be given. Then there is a $\delta\gt 0$ such that $\|T(x)-T(x_0)\|\lt \varepsilon$ whenever $\|x-x_0\|\lt \delta.$ Let $x_1$ be any arbitrary point in $X$. Then for $\|x-x_1\| = \|(x + x_0 -x_1)-x_0\|\lt \delta$, \begin{align*} \|T(x)-T(x_1)\| & = \|T(x)+T(x_0)-T(x_1)-T(x_0)\| \\ & = \|T(x + x_0 -x_1)-T(x_0)\|\qquad(\because T ~\text{is linear})\\ & \lt \varepsilon. \end{align*} (b)
$(\Rightarrow)$ Suppose $T$ is continuous. Then $T$ is bounded. So there exists an $M\geqslant 0$ such that for $x,y\in X$ $\|T(x)-T(y)\| \leq M \|x-y\|,$ which is precisely the condition for $T$ to be Lipschitz.
$(\Leftarrow)$ Suppose $T$ is Lipschitz. Then $T$ is bounded. Hence $T$ is continuous.
(c) at the moment, I can only think of $T(x) =x^2$ whichs is not linear but continuous everywhere. However, I can't quite deduce which of (a) or (b) fails to hold.