Problem Statement: If a real valued function on $\mathbb R^2$ is locally varying (on any non-empty open subset $U \subset \mathbb R^2$, the function is not constant), show that $\mathbb R^2$ cannot be written as
$$\mathbb R^2 = \bigcup^\infty_{i=1}\bigcup^\infty_{j=1}\{\,x : f_i(x) = c_j\,\}$$
where each $f_i \colon \mathbb R^2 \to \mathbb R$ is continuous and locally varying and each $c_j$ is a real number.
My attempt at a solution: I'm fairly certain that this is a Baire category theorem. I attempt to solve this by proving that for any $i,j \in \mathbb N$, $A_{ij} = \{\,x : f_i(x) = c_j\,\}$ is nowhere dense. Since $\mathbb R^2$ is complete, this should imply that we cannot write $\mathbb R^2$ as a countable union of these sets. However, I haven't done a lot of Baire Category theorem problems, and I'm getting a little tripped up in the reasoning. Here is what I have:
Assume for a contradiction that there is some $y \in \left(\overline{A_{ij}}\right)^o$. Now, this implies that for some $\epsilon > 0$, the open ball centered at $y$ with radius $\epsilon$, $B_\epsilon(y)$, is in $\overline{A_{ij}}$. Since $B_\epsilon(y)$ is open, $f_i$ is not constant on $B_\epsilon(y)$, but since $f_i(x) = c_j$ on $A_{ij}$, this means that for some $z \in B_\epsilon(y)$, $z$ is a point of closure of $A_{ij}$, but is not in $A_{ij}$. Thus, $f_i(z) = \alpha$, where $\alpha \neq c_j$.
Now, $f_i$ is continuous. Since $z \in \overline{A_{ij}}$, there exists some sequence of points in $A_{ij}$, say, $\{x_n\}_{n=1}^\infty$, such that $x_n \to z$. Thus, $f_i(x_n) \to f_i(z)$ as $n\to\infty$. But $f_i(x_n) = c_j \neq \alpha$, so $\lim_{n\to\infty} f_i(x_n) \neq f_i(z)$, a contradiction.
Thus, we have found that $A_{ij}$ has an empty interior for each $i,j \in \mathbb N$, so $\mathbb R^2$ cannot be written as a countable union of these sets, as desired.
So, that's what I have, but this seems way too simple, and I'm worried I've done something wrong! I'm mostly worried about the sequence of points $x_n \to z$ that I have set up - do I need this sequence to be in $B_\epsilon(y)$ as well, to get my contradiction?
You can use a shortcut: As $f_i$ is continous, the preimage $A_{ij}$ of the closed set $\{c_j\}$ is closed. As $f_i$ is locally varying, $A_{ij}$ cannot contain a nonempty open set. Hence $\left(\overline{A_{ij}}\right)^o=A_{ij}^o=\emptyset$.