Measurable functions in a countable co-countable $\sigma$-algebra

Let $I$ denote the image of $f$. If $I$ is not countable then some $r\in\mathbb R$ will exist such that $I\cap(-\infty,r)$ and $I\cap[r,\infty)$ are both not countable. Then the disjoint preimages of these sets cannot be countable, hence $f$ cannot be meausurable.

We conclude that $I$ must be countable. Then for $x\in I$ fibres $f^{-1}(\{x\})$ form a countable partition of $X=\mathbb R$, so at least one of these fibres is not countable. If for distinct $x,y\in I$ the fibres $f^{-1}(\{x\})$ and $f^{-1}(\{y\})$ are both uncountable then we can choose some $r$ in between $x$ and $y$ such that the disjoint preimages of $(-\infty,r)$ and $(r,\infty)$ are both uncountable, and again $f$ cannot be measurable.

We conclude that there is exactly one $x\in I$ with an uncountable fibre. The complement of this fibre is covered by other fibers. Each of them is a countable set and also the number of these fibers is countable. Conclusion: the complement of the mentioned uncountable fiber is countable, wich means that the uncountable fiber is cocountable.

Our final conclusion: $$f\text{ is measurable if and only if }f\text{ is constant on a cocountable set}$$ In that case the preimages of $(-\infty,r)$ with $r$ ranging over $\mathbb R$ will all be cocountable or countable.

edit (concerning question of @dan in a comment on this answer)

Let it be that $I$ is an uncountable subset of $\mathbb R$.

Let $A:=\left\{ x\in\mathbb{R}\mid\left(-\infty,x\right)\cap I\text{ is countable}\right\} $ and let $B:=\left\{ x\in\mathbb{R}\mid\left(x,\infty\right)\cap I\text{ is countable}\right\} $.

Note that the fact that $I$ is uncountable implies that $A\cap B=\varnothing$.

It is our aim to prove that $A\cup B\neq\mathbb{R}$.

So we assume that $A\cup B=\mathbb{R}$ and from here it is enough to find a contradiction.

At first hand for the shape of $A$ we see three possibilities: $A=\varnothing$, $A=\left(-\infty,s\right]$ for some $s\in\mathbb{R}$ or $A=\mathbb{R}$.

But if $A=\mathbb{R}$ then $I=\bigcup_{n=1}^{\infty}\left(I\cap\left(-\infty,n\right)\right)$ is countable as well, so the third possibility falls off.

Then similarly for the shape of $B$ we find two possibilities: $B=\varnothing$ or $B=\left[i,\infty\right)$ for some $i\in\mathbb{R}$.

Then based on $A\cup B=\mathbb{R}$ we find that also the possibilities $A=\varnothing$ and $B=\varnothing$ fall off.

So $\mathbb{R}=A\cup B=\left(-\infty,s\right]\cup\left[i,\infty\right)$ implying that $i\leq s$.

But then $A\cap B=\left[i,s\right]\neq\varnothing$ and a contradiction is found.

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Measure Theory

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