My question occurs when I see the problem that show that every continuous function is borel measurable. I know that for topological space $(X,\tau)$, we define that "open sets" means sets in topology. However, it is somewhat weird for me in this example;
Let $\tau$ be trivial (indiscrete) topology, which means that $\tau = \{X, \emptyset \}$. Let $X = \mathbb{R}$. Then only open set in this topological space is $\mathbb{R}, \emptyset$. However, we already know that any interval $(a,b)$ where $a,b \in \mathbb{R}$ is open!
So let's see the example of continuous function $f : (X,\tau) \to (X, 2^{X})$. In this case, does it really matter that any preimage of open sets in $2^{X}$ except $\emptyset$ is $X$ itself? I think this is intuitively confusing and somewhat contradiction occurs.
Am I thinking wrong or missing something on definition of topology?
Open intervals in $\mathbb{R}$ are not always open. What you are thinking about is what is called the usual topology on $\mathbb{R}$. This topology is defined to be the one generated by open balls, i.e. taking $B_\delta(x)=\{a\in\mathbb{R}|d(x,a)<\delta\}$ and then carrying out intersections and unions to discover what other sets must be open. If we have a different topology, then different sets are open.
The example of the topology you mention of just having $\{\},X$ be open is called the indiscrete topology. The discrete topology is when every set is open.