The strategy is a cyclic implication: (i) $\Rightarrow$ (ii) is trivial, (ii) $\Rightarrow$ (iii) uses the openness of inverse images, and (iii) $\Rightarrow$ (i) follows from the Lipschitz property of bounded [linear maps](/page/Linear%20Map). **Step 1: (i) $\Rightarrow$ (ii).** If $T$ is continuous at every point, then in particular it is continuous at $0$. **Step 2: (ii) $\Rightarrow$ (iii).** Suppose $T$ is continuous at $0$. Since $T(0) = 0$ (by linearity), the open unit ball $B_1^Y(0) \subset Y$ is an open neighbourhood of $T(0)$. By [continuity](/page/Continuity) at $0$, the preimage $T^{-1}(B_1^Y(0))$ is an open neighbourhood of $0$ in $X$. Therefore there exists $\varepsilon > 0$ such that $B_\varepsilon^X(0) \subset T^{-1}(B_1^Y(0))$, i.e. $T(B_\varepsilon^X(0)) \subset B_1^Y(0)$. Rescaling by $\varepsilon$: for any $x \in X$ with $\|x\|_X \le 1$, the element $\varepsilon x$ satisfies $\|\varepsilon x\|_X \le \varepsilon$, so $T(\varepsilon x) \in B_1^Y(0)$. By linearity, $\varepsilon T(x) = T(\varepsilon x)$, and hence $\|T(x)\|_Y < 1/\varepsilon$ for all $x$ with $\|x\|_X \le 1$. Setting $C = 1/\varepsilon$, we obtain $\|T(x)\|_Y \le C\|x\|_X$ for all $x \in X$ (the case $x = 0$ is trivial, and for $x \ne 0$ we apply the bound to $x/\|x\|_X$). **Step 3: (iii) $\Rightarrow$ (i).** Suppose $\|T(x)\|_Y \le C\|x\|_X$ for all $x \in X$. Then for any $x_1, x_2 \in X$: \begin{align*} \|T(x_1) - T(x_2)\|_Y = \|T(x_1 - x_2)\|_Y \le C\|x_1 - x_2\|_X. \end{align*} This shows $T$ is Lipschitz continuous with constant $C$, and Lipschitz maps are continuous everywhere.