Take as a model the concept of metric space, that is, a set $X$ endowed with a metric $d$. In this case, we define open set as a set $A\subset X$ such that$$(\forall a\in A)\bigl(\exists r\in(0,+\infty)\bigr):B_r(a)\subset A.\tag{1}$$It turns out that any arbitrary union of open sets is again an open set, because if $(1)$ holds for every element $A$ of a family $\mathcal F$ of open sets, then$$\left(\forall a\in\bigcup_{A\in\mathcal F}A\right)\bigl(\exists r\in(0,+\infty)\bigr):B_r(a)\subset\bigcup_{A\in\mathcal F}A;$$if $a\in\bigcup_{A\in\mathcal F}A$, you just take some $A\in\mathcal F$ such that $a\in A$, you take some $r\in(0,+\infty)$ such that $B_r(a)\subset A$, and then $B_r(a)\subset\bigcup_{A\in\mathcal F}A$.
If $\mathcal F$ is finite, then $\bigcap_{A\in\mathcal F}A$ is also an open set. In this casa, if $a\in\bigcap_{A\in\mathcal F}A$ you take, for each $A\in\mathcal F$, a $r_A\in(0,+\infty)$ such that $B_{r_A}(a)\subset A$ and you define $r=\min\{r_A\,|\,A\in\mathcal F\}$. Then$$B_r(a)=\bigcap_{A\in\mathcal F}B_{r_A}(a)\subset\bigcap_{A\in\mathcal F}A.$$
So, in metric spaces, an arbitrary union of open sets is open, but I only proved that a finite intersection of open sets must be open. And, in general, an arbitry intersection of open sets is not open. For instance, if $x\in X$, then $\{x\}=\bigcap_{n\in\mathbb N}B_{1/n}(x)$ but, in general, $\{x\}$ is not an open set (of course, it can be in some cases).
Then we define a topological space as a set $X$ endowed with a topology $\tau\subset\mathcal{P}(X)$, which, by definition, is closed with respect to arbitrary unions and finite intersections. That is, we take as a role model the situation that is familiar to us in the context of metric spaces.