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?

$f(x_0+h)=f(x_0)f(h)$, as $f(h)\to f(0)$ ($h\to 0$), then $f(x_0+h)=f(x_0)f(h)\to f(x_0)f(0)=f(x_0)$.
