real analysis - Continuity in dense set

Prove that if $f$ is continuous and $f(x)=0$ for all numbers $x$ in a dense set $A$, then $f(x)=0$, for all $x$ .

Proof. Suppose that $f(x) \neq 0$, for some $x$, as $f$ is continuous then there is $\delta>0$, such that $(x-\delta,x+\delta)$, given that a set $A$ is dense, there is $\alpha \in (x-\delta,x+\delta)$, for $\alpha \in A$, then $f(\alpha)=0$, but this is a contradiction, since $\alpha \in A$, hence $f(x)=0$ .

I'm not sure if it's necessary to use the case in which $f(x)<0$ and $f(x)>0$ .

Other Proof:

Given that A is a dense set, there is $x_{n}\in A$, such that $x_{n}\rightarrow x$, then $f$ is continuous $f(x_{n})\rightarrow f(x)$, given that $x_{n} \in A$, then $f(x_{n})=0$, thus $f(x)=0$