Continuity of function where $f(x+y) = f(x)f(y) ~~\forall x, y \in \mathbb{R}$. Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be a function which satisfies $f(x+y) = f(x)f(y)  ~~\forall x, y \in \mathbb{R}$ is continuous at $x=0$, then it is continuous at every point of $\mathbb{R}$.
So we know $\forall \epsilon > 0 ~~\exists \delta > 0$ such that $|x-0|<\delta \implies |f(x)-f(0)|<\epsilon$ and we want to show that given any $\epsilon > 0 ~~\exists \delta > 0$ such that $|x-y|<\delta \implies |f(x)-f(y)|<\epsilon$.
Now I see that the 'trick' that we can use is replacing $f(x)$ with $f(x)f(0)$ but I still cannot seem to finish the proof. Any advice?

To show continuity at $x$ simply notice that:
$$\lvert f(x+h)-f(x)\rvert=\lvert f(x)f(h)-f(x)\rvert=\lvert f(x)\rvert\lvert f(h)-1\rvert$$
and notice that $f(0)=f(0+0)=f(0)f(0)$ so either $f(0)=0$ or $f(0)=1$. So either $f$ is identically $0$ and hence continuous or we use continuity at $x=0$.
