Proof of Restricting a Continuous Map to a Smaller Subset Preserves Continuity

Since $B\subseteq A$, every $y\in B$ lies in $A$ and satisfies $f|_{B}(y)=f(y)$ (Domains and Restrictions of Maps (iii)).

Proof of (i). Let $x\in B$, so that $x\in A$ as $B\subseteq A$, suppose that $f$ is continuous at $x$ relative to $A$, and let $\varepsilon$ be a real number with $\varepsilon>0$. By Continuous Map Between Metric Spaces (i) there is a real $\delta>0$ such that every $y\in A$ with $d_X(x,y)<\delta$ satisfies $d_Y(f(y),f(x))<\varepsilon$. Let $y\in B$ with $d_X(x,y)<\delta$. Then $y\in A$, so $d_Y(f(y),f(x))<\varepsilon$; and as $x,y\in B$, $f|_{B}(y)=f(y)$ and $f|_{B}(x)=f(x)$, so

$$d_Y\bigl(f|_{B}(y),f|_{B}(x)\bigr)=d_Y(f(y),f(x))<\varepsilon .$$

As $\varepsilon>0$ was arbitrary, Continuous Map Between Metric Spaces (i), applied to the subset $B$ of $X$ and the map $f|_{B}:B\to Y$, shows that $f|_{B}$ is continuous at $x$ relative to $B$.

Proof of (ii). Suppose that $f$ is continuous on $A$, and let $x'\in B$. Then $x'\in A$, so $f$ is continuous at $x'$ relative to $A$ by Continuous Map Between Metric Spaces (ii), and (i), applied to $x'$, shows that $f|_{B}$ is continuous at $x'$ relative to $B$. As $x'\in B$ was arbitrary, $f|_{B}$ is continuous on $B$ by Continuous Map Between Metric Spaces (ii).