Interior of a preimage of a continuous function 
Let $ f:\mathbb{R}^n\rightarrow \mathbb{R} $ be convex.
Let there exist a point $ x_0 $ with $ f(x_0)<0 $.
Prove that
$$ \operatorname{int}\left\lbrace f(x)\ge 0 \right\rbrace = \left\lbrace f(x)> 0 \right\rbrace. $$

Where "int" denotes the interior of the set? The inclusion from the right to the left is clear to me, but the inclusion from left to right $ \subseteq $ is not.
Perhaps, it could start like this:

Let $ x \in \operatorname{int}\left\lbrace f(x)\ge 0 \right\rbrace $ be given.
Assume $ x \notin\left\lbrace f(x)> 0 \right\rbrace $, then $ f(x)=0 $ has to hold.
Since $ x \in \operatorname{int}\left\lbrace f(x)\ge 0 \right\rbrace $ there exists a ball $ B(x,\varepsilon) \subseteq \operatorname{int}\left\lbrace f(x)\ge 0 \right\rbrace $.

And, as a supplementary question: Is the existence of $ x_0 $ with $ f(x_0)<0 $ necessary?

Your start is good, if you're going for a proof by contradiction. So, suppose you have an $x$ with $f(x)=0$ such that $B(x,\varepsilon)\subseteq\{f(x)\geq 0\}$. Consider what happens along the line segment connect $x$ to some $v$ such that $f(v)<0$. In particular, for every $\alpha\in (0,1]$ we get that $f(\alpha v + (1-\alpha) x)<0$ by convexity. Now, just choose $\alpha$ small enough so that $\alpha v + (1-\alpha)x \in B(x,\varepsilon)$, and you've cooked up your contradiction!
The condition that there be a place where the function is negative is necessary; the statement is clearly false for the function $f(x)=0$ or $f(x)=x^2$.
