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?
 A: Since the function is convex, let $x$ be an element of the interior $U_0$ of $C_0=\{x\in R^n, f(x)\leq 0\}$.Suppose that $f(x)=0$, there exists an open ball $U\subset U_0$ which contains $x$, $f(U_x)$ is a connected interval since $f$ is continue.
Suppose that for every $x\in U_0$ such that $f(x)=0$, $f(U_x)$ is the singleton $\{0\}$, this implies that $D=\{x,:U\in U_0, f(x)=0\}$ is open and closed, it is empty or equal to $R^n$, since there exists $x_0$ such that $f(x_0)<0$, $D$ is empty.
Suppose that $f(U_x)=[c,0]$ this is impossible since $f$ is convex. (You can find an open interval $I\subset U$ such that $f(I)=[d,0], d<0$).
We deduce that the image of an element in the interior of $C_0$ cannot be equal to zero.
This is not true,if the function is not convex consider $f:[-1,1]\rightarrow R$ defined by $f(x)=x, x\leq 0$, $f(x)=-x, x>0$, $f$ is continuous, and $f([-1,1])=[-1,0]$.
$\{x,f(x)<0=[-1,1]-\{0\}$ is not the interior of $[-1,1]$.
A: 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 $f(x)=x^2$.
