Proof of A Uniformly Continuous Map Between Metric Spaces Is Continuous

Each result cited is universally quantified over the data in its own statement.

Let $x\in A$ and let $\varepsilon\in\mathbb{R}$ be positive. By Uniformly Continuous Map Between Metric Spaces there is a positive $\delta\in\mathbb{R}$ such that all $z,z'\in A$ with $d_{X}(z,z')<\delta$ satisfy $d_{Y}(f(z),f(z'))<\varepsilon$.

Let $y\in A$ satisfy $d_{X}(x,y)<\delta$. Taking $z=x$ and $z'=y$, which are admissible since $x,y\in A$, gives $d_{Y}(f(x),f(y))<\varepsilon$. By condition 3 of Metric Space applied to the metric $d_{Y}$, $d_{Y}(f(y),f(x))=d_{Y}(f(x),f(y))$, so $d_{Y}(f(y),f(x))<\varepsilon$.

Thus for every positive $\varepsilon$ there is a positive $\delta$ such that every $y\in A$ with $d_{X}(x,y)<\delta$ satisfies $d_{Y}(f(y),f(x))<\varepsilon$; that is, $f$ is continuous at $x$ relative to $A$ by Continuous Map Between Metric Spaces. As $x$ was an arbitrary point of $A$, $f$ is continuous on $A$ by that same item.