Let $f(x)$ be a continuous function on $\Bbb{R}$ and $a \in \Bbb{R}$
$\forall \epsilon>0$ we define the set $$S_{\epsilon}=\{\delta>0|\forall x:0<|x-a|<\delta \Rightarrow 0<|f(x)-f(a)|<\epsilon\}$$
Assume that $\forall \epsilon>0$ we have $S_{\epsilon}=(0,+\infty)$
What can someone conclude about $f(x)$?
I believe that $f$ is constant $f(a)$
Assume that $\exists x_0 \in \Bbb{R}$ such that $|f(x_0)-f(a)|>0$
Put $\epsilon_0=|f(x_0)-f(a)|>0$
Then we have that $S_{\epsilon_0}=(0,+\infty)$
We choose $\delta_0>0$ big enough such that $x_0 \in (a-\delta_0,a+\delta_0)$
Then we can easily see that $\delta_0 \notin S_{\epsilon_0}\Rightarrow S_{\epsilon_0} \neq (0,+\infty)$ which is a contradiction.
Thus $f(x)=f(a), \forall x \in \Bbb{R}$.
Is my proof correct?
Thank you in advance.