More generally, can the zero set $V(f)$ of a continuous function $f : \mathbb{R} \to \mathbb{R}$ be nowhere dense and uncountable? What if $f$ is smooth?
Some days ago I discovered that in this proof I am working on, I have implicitly assumed that $V(f)$ has to be countable if it is nowhere dense - hence this question.
Here's a semi-explicit construction for a smooth function f that is zero precisely on the classical Cantor set. By this set I mean the one that is obtained from $I_0 = [0,1]$ by repeatedly removing the middle third of any ensuing interval. So let's denote by $I_n$ the $n$-th set in this process.
Now let's make a smooth function $f_n$ on $[0,1]$ such that its zero set is exactly $I_n$. Starting with $f_0 = 0$ we obtain $f_{n+1}$ from $f_n$ as follows:
Set $f_{n+1} = f_n$ on $I_{n+1}$ and on an interval that is removed from $I_n$ make $f_{n+1}$ equal to a bump function that is 0 only at the boundary of the interval. We can choose the bump function to be of height $2^{-2^n}$.
This choice of heights of the bump functions will ensure that the derivatives of $f$ all converge uniformly to their pointwise limits. Hence the limit function $f_n$ is again smooth. By construction its zero set is exactly the Cantor set.
The continuous function is very easy to construct: it's the distance to the closed set.
It is a standard result that each closed subset of $\mathbb{R}^n$ is a zero set of some smooth function on $\mathbb{R}^n$. One proves this using smooth bump functions and partitions of unity.
In a normal topological space, the zero-sets of continuous functions are precisely the closed $G_{\delta}$ sets. Hence in any metric space all closed sets are, including the Cantor set.
I can't resist trying my hand at sketching a proof of the general result given in Robin Chapman's answer. Let $F\subset\mathbb{R}^n$ be any closed set. Let $E_0=\{x\colon\operatorname{dist}(x,F)\ge1\}$, and for $k=1,2,\ldots$ let $E_k=\{x\colon2^{-k}\le\operatorname{dist}(x,F)\le2^{1-k}\}$. Let $\omega$ be a standard mollifier, and put $$f=\sum_{k=0}^\infty \alpha_k\chi_{E_k}*\omega_k,\qquad\omega_k(y)=2^{nk}\omega(2^ky),$$ where $\alpha_k>0$ decays fast enough so all derivatives converge uniformly ($\alpha_k=2^{-k^2}$ ought to be sufficient).
Here's an answer from probability: a Brownian motion $B_t$ is a random, continuous function whose zero set is closed, nowhere dense, and has no isolated points. That is, $\{t : B_t = 0 \}$ is almost-surely a topological Cantor set (see, for example, Section 8 of Lalley's lecture notes).
To mention a further point not covered in existing answers: while any closed subset of $\mathbb{R}$ can be the zero set of a smooth function, the zero set of an analytic function either consists entirely of isolated points, or is all of $\mathbb{R}$. To see this we note that if the zero set of an analytic function $f$ contains an accumulation point, then by taking a power series expansion of $f$ at the accumulation point we may extend $f$ locally to a small complex disc around that point, and apply the Identity Theorem from complex analysis to show that $f$ is everywhere zero within that disc. In particular the zero set contains an open neighbourhood in $\mathbb{R}$ of the accumulation point, and using connectedness we can repeat this argument to show that the zero set must be all of $\mathbb{R}$.
André Henriques answer for a closed set $C$ can easily be improved to $C^\infty$ by considering $e^{1/(\alpha-x)+1/(x-\beta)}$ if $x\not\in C$ where $\alpha$ is the supremum of all elements $< x$ in $C$ (and $\alpha=-\infty$ if $C$ contains no elements which are $< x$) and where similarly $\beta$ is the infimum of all elements $> x$ in $C$ (respectively $\beta=\infty$ if $C$ contains no elements $> x$).
