real analysis - A nontrivial everywhere continuous function with uncountably many roots?

This is my first post on SE, forgive any blunders.

I am looking for an example of a function $f:\mathbb{R} \to \mathbb{R}$ which is continuous everywhere but has uncountably many roots ($x$ such that $f(x) = 0$). I am not looking for trivial examples such as $f = 0$ for all $x$. This is not a homework problem. I'd prefer a nudge in the right direction rather than an explicit example.

Thanks!

Edit: Thanks all! I've constructed my example with your help.

Answer

The roots of a continuous function is always a closed subset of $\mathbb{R}$ : $\{0\}$ is closed, thus $f^{-1}(\{0\})$ is closed too.

If you have a closed set $S$, you can define a function $f : x \mapsto d(x,S)$, which is continuous and whose set of roots is exactly $S$ : you can make a continuous function have any closed set as its set of roots.

Therefore you only have to look for closed sets that are uncountable.