In a normed vector space, if $S⊆X$ is closed, is $f(S,r)$ necessarily closed?

Question

Let $X$ denote a (real or complex) normed vector space.

Consider the function $f : \mathcal{P}(X) \times \mathbb{R}_{\geq 0} \rightarrow \mathcal{P}(X)$ given as follows. $$f(S,r) = \bigcup_{s \in S}\overline{B(s,r)}.$$

We know from William's answer here that if $S \subseteq X$ is compact, then $f(S,r)$ is closed.

Question. If $S \subseteq X$ is merely assumed closed, is $f(S,r) \subseteq X$ necessarily closed?

This is not the case for $X$ is an arbitrary metric space; see Bryan's answer (here's that link again) for a counterexample.

Answer 1

No, $f(S,r)$ need not be closed for closed $S$.

Let $X = \ell^p(\mathbb{N})$ and $S = \left\{ \left(1+\frac{1}{n+1}\right)\cdot e_n : n \in \mathbb{N}\right\}$. $S$ is closed, but $0 \in \overline{f(S,1)} \setminus f(S,1)$.