Is the set of isolated zeros of a continuous function closed?

Question

If $f:\mathbb R\to\mathbb R$ is continuous, is $$B:=\left\{x\in\mathbb R:x\text{ is an isolated point of }f^{-1}\left(\left\{0\right\}\right)\right\}$$ closed? It's clear that $B$ is countable, but since a countable set doesn't to be closed ($\left\{\frac1n:n\in\mathbb N\right\}$ being a counterexample), that doesn't yield the claim. (As usual, I'm assming $\mathbb R$ is equipped with the euclidean topology.)

Answer 1

No. For instance, let $A\subseteq \mathbb{R}$ be any closed set and let $f(x)=d(x,A)$. Then $f^{-1}(\{0\})=A$. But, the isolated points of $A$ do not need to be closed; for instance, $A$ could be $\{0\}\cup\{1/n:n\in\mathbb{Z}_+\}$.

Answer 2

No. Take, for instance$$\begin{array}{rccc}f\colon&\mathbb R&\longrightarrow&\mathbb R\\&x&\mapsto&\begin{cases}x\sin\left(\frac1x\right)&\text{ if }x\neq0\\0&\text{ otherwise.}\end{cases}\end{array}$$Then the set of isolated zeros is $\left\{\frac1{\pi n}\,\middle|\,n\in\mathbb Z\setminus\{0\}\right\}$, which is not closed ($0$ belongs to its closure, but not to the set itslef).