Proof of Continuity on a Subset Is Continuity on the Metric Subspace

Proof of (i). Let $x\in A$. By Continuous Map Between Metric Spaces (i), applied to $(X,d)$ and the subset $A$, $f$ is continuous at $x$ relative to $A$ if and only if the following condition $(\ast)$ holds: for every real $\varepsilon>0$ there is a real $\delta>0$ such that every $y\in A$ with $d(x,y)<\delta$ satisfies $e(f(y),f(x))<\varepsilon$. Applied to the metric space $(A,d_{A})$ and its subset $A$ itself, the same definition says that $f$ is continuous at $x$ as a function from $(A,d_{A})$ if and only if $(\ast)$ holds with $d_{A}(x,y)$ in place of $d(x,y)$. For $y\in A$, $d_{A}(x,y)=d(x,y)$, as $(x,y)\in A\times A$ (Domains and Restrictions of Maps (iii)); so the two conditions are the same, and the two notions agree.

Proof of (ii). By Continuous Map Between Metric Spaces (ii), $f$ is continuous on $A$, as a function from a subset of $(X,d)$, if and only if it is continuous at every $x'\in A$ relative to $A$; and $f$ is continuous as a function from $(A,d_{A})$ if and only if it is continuous at every $x'\in A$ relative to the whole space $A$ of $(A,d_{A})$. By (i), applied at each $x'\in A$, these two conditions agree.