In what sense do the zeroes of continuous functions vary continuously? Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function and $r \in \mathbb{R}$ such that $f(r) = 0$ and furthermore $f$ crosses the $x$-axis at $r$.  Intuitively, if you change $f$ slightly, the zero at $r$ shouldn't disappear and should only change slightly.  How can this intuition be made precise?  What sort of metric or topology do we need to put on $C(\mathbb{R},\mathbb{R})$?
 A: I think this is a delicate matter. I am not sure if you will be able to make the intuition precise requiring only continuity. 
For instance, one attempt would be to consider the continuous bounded functions together with the uniform norm
$$\Vert f\Vert_{0}= \sup_{x \in \mathbb{R}} |f(x)|.$$
This seems to align with the "small changes" idea. The problem is that no matter how small you vary the function in the above norm,  you can have an arbitrary amount of wiggling near the zero. In particular, you can have an arbitrary amount of new zeroes. It may be difficult to keep track of your original zero, whatever that would mean.
However, it is indeed possible if you assume $f$ to be $C^1$. Your condition that $f$ crosses the $x$-axis at $r$ can then be translated to $f'(r) \neq 0$. If we now consider the $C^1$-norm (i.e., $\Vert f\Vert_1:=\Vert f\Vert_{0}+\Vert f'\Vert_1$) on the space of bounded $C^1$ functions with bounded derivative, we then have a complete space $C^1_B(\mathbb{R})$. The map
$$G: C^1_B(\mathbb{R}) \times \mathbb{R} \to \mathbb{R}$$
$$(f,x) \mapsto f(x)$$
then gives a $C^1$ function. By the implicit function theorem (note that we are using that $f'(r) \neq 0!$), in a small neighbourhood of a given $f$ such that $f(r)=0$ in the $C^1_B$ topology there will be a continuous (in fact, $C^1$) association $g \mapsto r_g$, where $r_g$ is such that $g(r_g)=0$.
Some pictures to illustrate my point (the first being a function close to the original in the $C^0$ case and the second in the $C^1$ case):
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