Lets define a function $z:\mathbb{R}^\mathbb{R}\to\mathcal P(\mathbb R)$ that gives you the set of zeros of any $\mathbb R ^\mathbb R$ function. Now, we define a set $S=\{z(f):f\in\mathbb R ^\mathbb R, f \text{ continuous}, f\neq0\}$.
The question is, what is the cardinality of the largest element of $S$.
In other words, avoiding the above which is probably just an exercise in LaTeX, what is the maximum number of zeros that a continuous function can have?
I think the answer is $\aleph_0$ because we can find a unique rational number between any two zeros, so there can't be more of those intervals than there are rational numbers and the number of intervals is the same as the number of zeros.
But honestly, I don't know why the same argument can't be applied to a function that is $0$ for every irrational number and $1$ for every rational.
Let $f(x) = \begin{cases} 0 & x \le 0 \\ x & x > 0 \end{cases}$.
$f$ is continuous and is $0$ for $x \le 0$, which has cardinality $\mathfrak c$.
Even if we specify our function to be $C^{\infty}$ (infinitely differentiable), we can still have uncountably many zeroes. Consider:
$$g(x) = \begin{cases} 0 & x \le 0 \\ e^{-1/x^2} & x > 0 \end{cases}$$
OK, what if we specified our function to be entire, that is, complex differentiable? Then we can only have at most countably many zeroes (as is the case with $\sin z$) since the zeroes of a nonzero complex differentiable function must be discrete, you can't have a limit point of zeroes.
OK, what if we specified that our function is never $0$ over an interval and $C^{\infty}$? This would require the set of zeroes to be nowhere dense. We can still construct a function with uncountably many zeroes (take a sum of bump functions of height $1/2^n$ in the gaps of the Cantor set deleted at the $n$-th step of construction).