Distance is (uniformly) continuous

The reason $d(x,A)$ is a uniformly continuous function is that $|d(x,A) - d(y,A)| \le |x -y|$ for all $x$ and $y$. Let $x$ and $y$ be given. For every $z\in A$, $$d(x,A) \le |x-z| \le |x-y| + |y-z|.$$ Thus $d(x,A) - |x-y|$ is a lower bound for the set $\{|y-z|:z\in A\}$. Hence $d(x,A) - |x - y| \le d(y,A)$, i.e., $d(x,A) - d(y,A) \le |x - y|$. A similar argument shows that $d(y,A) - d(x,A) \le |x - y|$. Hence $|d(x,A) - d(y,A)| \le |x - y|$.

$$ d(w,y)\le d(w,x)+d(x,y) $$ $$ \inf_{y\in A} d(w,y) \overset{\text{?}} \le \inf_{y\in A}(d(w,x)+d(x,y)) \overset{\text{?}}= d(w,x)+\inf_{y\in A} d(x,y) $$

Can you then show that if $d(w,x)<\varepsilon$ then $|d(w,A)-d(x,A)|<\varepsilon$?