real analysis - Is the zero set of a function open if the function is continuous on $mathbb R$?

f: $\mathbb R \to \mathbb R$ is continuous on $\mathbb R$. S:={x $\in\mathbb R$, f(x) = 0}. Is S open if the function is continuous on $\mathbb R$?

I tried to pick up an arbitrary point $\alpha$ in S.
Since the function is continuous on $\mathbb R$.

I get $\forall \varepsilon >0, \exists \delta > 0$, such that | f (x) - f ($\alpha$) | < $\varepsilon$, $\forall x \in V \delta (\alpha)$.

Since f($\alpha$)=0, 0 $\le| f(x) |< \varepsilon$.
Then f(x) = 0, x $\in$ S, $\forall x \in V \delta (\alpha)$.

Thus $\exists \delta >0, V \delta (\alpha) \subseteq S , \forall \alpha \in S$, which means S is open.

Is this well-proved or is there anything wrong?