Let $f_n^+(x)$ be a sequence of non-negative functions $f_n^+: X \to \Bbb{R}_{\geq 0}$, such that each $f_n^+$ has countably many zeros. Then if $f(x) = \sum f_n^+(x)$ converges point-wise, the number of zeroes of $f$ is countable. Proof: $f^+(x) = 0$ iff $f_n^+(x) = 0, \ \forall n \implies Z(f) = \bigcap Z(f_n^+)$, which is clearly countable. The converse doesn't seem necc. true.
Obviously we can't say the same about any other level sets of $f^+$. Imagine the level sets of a continuous mountain. Each topographical ring around the mountain has size equal to a closed interval of $\Bbb{R}$.
Is'nt that weird? I must be wrong then...