Proof of A Uniformly Continuous Map Between Metric Spaces Is Continuous

Let $x\in A$ and let $\varepsilon>0$ be real. By Uniformly Continuous Map Between Metric Spaces §uniformly-continuous there is a real $\delta>0$ such that all $z,z'\in A$ with $d_{X}(z,z')<\delta$ satisfy $d_{Y}(f(z),f(z'))<\varepsilon$. Let $y\in A$ with $d_{X}(x,y)<\delta$. Taking $z=x$ and $z'=y$ gives $d_{Y}(f(x),f(y))<\varepsilon$, and $d_{Y}(f(y),f(x))=d_{Y}(f(x),f(y))$ by Metric Space §symmetric, so $d_{Y}(f(y),f(x))<\varepsilon$. As $\varepsilon$ was arbitrary and $\delta$ was chosen after it, $f$ is continuous at $x$ relative to $A$; as $x\in A$ was arbitrary, $f$ is continuous on $A$ by Continuous Map Between Metric Spaces §continuous.