 Solution

The Bolzano-Weierstrass theorem in $\mathbb{R}$ says that every bounded real sequence has a convergent subsequence. For a bounded sequence in ${\mathbb{R}}^n$, successively extract subsequences on which the first, second, and so on through the $n$th coordinate converge. The final diagonal subsequence converges in Euclidean space, proving the theorem in ${\mathbb{R}}^n$.

Put $K=D\setminus {\bigcup }_{z\in S}{B}_{\epsilon }(z)$. Suppose the asserted $\delta$ did not exist. Then there would be $x_k\in D$ and $y_k\in K$ with $\Vert x_k-y_k\Vert <1/k$ but $|f(x_k)-f(y_k)|\ge \epsilon$. Since $K$ is closed and bounded, Bolzano-Weierstrass gives a subsequence $y_k\to y\in K$, and then $x_k\to y$. The point $y$ is not in $S$, because otherwise $y\in B_{\epsilon }(y)$, so $f$ has continuity at $y$. Both $f(x_k)$ and $f(y_k)$ tend to $f(y)$, a contradiction. Hence the required uniform $\delta >0$ exists.