We define $f: X \to \mathbb{R}$ such that for every open interval $(a,b)$ its preimage $f^{-1}\left((a,b)\right)$ is open.
Does it imply that f is continuous?
I know the opposite is not true (e.g for $f(x) = x^{2}$ we have $f^{-1}\left((1,4\right)) = (-2, -1) \cup (1,2) $ which is not an open interval (since any number between -1 and 1 is not there).
It looks like this question has already been answered here - Function in which every inverse image of open "interval" is open interval but not continuous. - but unfortunately I am unfamiliar with bases and could not get a good grasp of the answer.
Is there an easy way to prove it without that term?
Yes. Because preimages work nicely with set operations.
If you take any open set $V\subset \mathbb R$, you can write $V=\bigcup_j(a_j,b_j)$ (this is very easy to prove: for every point in $V$ there is an interval that contains the point and is contained in $V$). Thus $$ f^{-1}(V)=f^{-1}(\bigcup_j(a_j,b_j))=\bigcup_j f^{-1}((a_j,b_j)) $$ is a union of open sets so open. So $f$ is continuous.