I was wondering if my textbook's proof was superfluous. It makes use of closed balls which don't seem necessary.
"If (U_n) is a sequence of dense open subsets of S, then the intersection X = (intersection over all n) U_n is dense in S.
Consider a nonempty open subset V of S; we need to show X ∩ V = ∅. Since U_1 is open and dense and V is open, U_1 ∩ V is open and nonempty. So there exists x_1 ∈ U_1 ∩ V and r_1 ≤ 1 so that C_r_1 (x_1) ⊆ U_1 ∩ V . Since U_2 is open and dense and B_r_1 (x_1) is open, there exists x_2 ∈ B_r_1 (x_1) ∩ U_2 and r_2 ≤ 1/2 so that C_r_2 (x_2) ⊆ B_r_1 (x_1) ∩ U_2 . Continuing, we obtain sequences (x_k) in S and positive numbers (r_k) so that
x_(k+1) ∈ B_(r_k) (x_k), r_k ≤ 1/(2^(k-1)) and C_(r_(k+1)) (x_(k+1)) ⊆ B_(r_k) (x_k) ∩ U_(k+1)
... (omitting some details showing sequence is Cauchy for visual clarity)
Thus the sequence (x_k) is a Cauchy sequence in S. Since S is complete, we have lim x_k = x for some x ∈ S. Since we have
x_k ∈ C_(r_k) (x_k) ⊆ C_(r_n) (x_n) for k ≥ n,
and since C_(r_n) (x_n) is closed, x also belongs to C_(r_n). This is true for all n, so x is an element of intersection of C_(r_n)(x_n) over all n, which is a subset of the intersection of U_n over all n which equals X.
Since we also have x ∈ C_(r_1) (x_1) ⊂ V , we conclude x ∈ X ∩ V , as desired."
Doesn't the proof work just fine with only the nested open balls, since we know the sequence is Cauchy and can show that the limit must be contained in all the open balls?
No, because for example $n^{-1} \in (0,2n^{-1})$ for all $n\geq 1$, but the limit (namely $0$) is not contained in $\bigcap_1^\infty (0, 2n^{-1})$. The key to making this proof work in not only metric spaces but a more general topological setting (i.e., compact Hausdorff spaces) is that for every open $U$ and point $x\in U$ one can find a neighborhood $D$ containing $x$ such that $D \subset \overline D \subset U$.