Essential support vs. classical support for a continuous function

The essential support of a function $f:\Bbb R^n\rightarrow \Bbb R$ is defined in the following way:

Let's denote $\mathcal A_f=\{\omega \subset \Bbb R^n: \omega \quad \text{open}, \quad f(x)=0\quad \text{a.e.} \quad x\in \omega\}$ and $A_f=\cup_{\omega \in \mathcal A_f}\omega$. The essential support $supp_e f$ of $f$ is $\Bbb R^n -A_f$.

I want to show that, if $f$ is continuous, then $supp_e f=supp f$.

My attempt:

Now, let $\omega \in \mathcal A_f$. Then, there exists $N_\omega\subseteq \omega$ such that $|N_\omega|=0$, and $f(x)=0$ for each $x\in \omega- N_\omega$, and suppose that $N_\omega \neq \varnothing$. Let $x\in N_\omega$, so $f(x)\neq 0$. Since $f$ in continuous in $x$, there exists a ball $B$ centered in $x$, such that $f(y)\neq 0$ for each $y\in B$, so $B\subset N\omega$. Thus $|B|=0$, which is impossible. It follows that $N_\omega = \varnothing$. (Is it correct?)

Then $f(x)=0$ for each $x\in \omega$, for each $\omega\in \mathcal A_f$ $(*)$. From $(*)$ follows that $supp_ e f= supp f$.

In fact, if $x\in supp_e f$, then $x\notin A_f$, thus $f(x)\neq 0$, that is $x\in supp f$. On the other hand, suppose that $x\in supp f$. Then either $f(x)\neq 0$, or $x\in \overline{\{x\in \Bbb R^n:f(x)\neq0\}}-\{x\in \Bbb R^n:f(x)\neq0\}$. In the former case, we have that $x\notin A_f$, so $x\in supp_e f$. In the latter case, there are points of $\{x\in \Bbb R^n:f(x)\neq0\}$ arbitrarily closed to $x$, and again, since $f$ is continuous, $f(x)\neq 0$, so we are again in the previous case.

Is it correct? Or is there an easiest way to solve the problem?

Everything is good up to "In the latter case, ...". It may well be that $x$ is a limit point of $\{f \neq 0\}$ while $f(x) = 0$. The identity map on $\mathbb R$ at $0$ is an example of this.

To finish off the proof, all that's needed is that $\overline{\{f \neq 0\}}^c \in \mathcal A_f$. Clearly $\overline{\{f \neq 0\}}^c$ is open and since $\{f \neq 0\} \subset \overline{\{f \neq 0\}}$, we have that $\overline{\{f \neq 0\}}^c \subset \{f \neq 0\}^c = \{f = 0\}$.