Let $f: \Omega \to [0, \infty]$ be measurable. Let $\mu$ be $\sigma$-finite. Then $\{t\mid \mu\{f = t\} \neq 0 \}$ is countable. Let $f: \Omega \to [0, \infty]$ be measurable. Let $\mu$ be $\sigma$-finite. Then $\{t\mid \mu\{f = t\} \neq 0 \}$ is countable.
My attempt:
I managed to show the result if $\mu(\Omega) < \infty$.
If $\mu(\Omega) =\infty$, I guess I'll have to use $\sigma$-finiteness to lift the previous case.
I took a sequence $(C_n)_n$ with $\mu(C_n) < \infty$ such that $C_n \uparrow \Omega$.
I then defined $\mu_n(A) := \mu(A \cap C_n)$. Then $\mu_n$ is a $\sigma$-finite measure and $\{t \mid \mu_n\{f = t\} \neq 0\}$ is at most countable.
Observe now that $\mu = \lim_n \mu_n$. From here, I'm stuck.
Any ideas? (Note, the hypothesis that $\mu$ is $\sigma$-finite might be irrelevant)
 A: You are off to a good start. Let $t \in \mathbb R$ and let  $A_t = \{f = t\}$. Since $C_n \uparrow \Omega$ you have 
$$\mu(A_t) = \lim_{n \to\infty} \mu(A_t \cap C_n).$$
Since $\mu(A_t) > 0$ there exists an index $n(t)$ with the property that $\mu(A_t \cap C_{n(t)}) > 0$.
Regard $n$ as a function $\mathbb R \mapsto \mathbb N$. A basic cardinality argument tells you that there is at least one $n_0 \in \mathbb N$ with the property that $n^{-1}(n_0)$ is uncountable.
It follows that the family $\{A_t \cap C_{n_0}\}$ is an uncountable family of subsets of $C_{n_0}$ each having positive measure. Since $\mu(C_{n_0}) < \infty$ you can appeal to the finite measure case you already proved.
A: Assume that $\mu$ is a positive $\sigma$-finite measure.
Let $t \in [0,\infty]$ so that $\mu\{f=t\} > 0$.  Since OP has noticed that $\mu = \lim_n \mu_n$, there exists $N \in \Bbb{N}$ such that for all $n\ge N$, $\mu_n\{f=t\} > 0$.  Therefore,
$$\{t\mid \mu\{f = t\} >0 \} \subseteq \bigcup_N \bigcap_{n \ge N} \{t\mid \mu_n\{f = t\} > 0 \}.$$
OP has proved that for each $n$, $\{t\mid \mu_n\{f = t\} > 0 \}$ is countable, so as their countable intersection $$\bigcap_{n \ge N} \{t\mid \mu_n\{f = t\} > 0 \},$$
as well as the countable union
$$\bigcup_N \bigcap_{n \ge N} \{t\mid \mu_n\{f = t\} > 0 \}$$
of countable sets.
The case for signed measure follows the same logic, and I leave this for OP.
