Why is that *any* union of open sets is open but only *finitely many* intersections of open sets is open?

Generally this goes back to the underlying concept of convergence of a sequence and the so called Hausdorff'sche Umgebungsaxiome (see: Felix Hausdorff. Grundzüge der Mengenlehre. Veit & Comp., Leipzig, 1914).

Intuitively the convergence of a sequence to a point means, that in each neighbourhood of the point can be found almost all parts of the sequence.

The mathematical version of that concept is to name for each point $x$ of a set $X$ certain subsets as neighbourhoods of this point satisfying the following so-called Hausdorff'sche Umgebungsaxiome (neighbourhood = germ. "Umgebung"):

1. $x$ belongs to each of its neighbourhoods.
2. Any superset of a neighborhood of $x$ is again a neighborhood of $x$ (esp. $X$ is a neighborhood of $x$).
3. The intersection of two neighborhoods of $x$ is again a neighborhood of $x$.
4. Each neighborhood $U$ of $x$ contains a neighborhood $V$ of $x$ such that $U$ is a neighborhood of each point of $V$.

On this basis, one defines a subset of $X$ to be open if it is a neighborhood of each of its points (f.i. a circular disk in $R^2$ without its edge).

Open are exactly those subsets, taht contain only "internal" points (and neither "outer points" or "edge-points") in the sense that at any point in the open set a neighbourhood can be specified, which is completely contained in the open set.

From this can be derived following theorems:

1. The empty set and the space itself are open.
2. The intersection of two open sets is open.
3. The union of any number of open sets is open.
4. A subset $U$ is exactly then a neighborhood of $x$ if there exists an open set $O$ with $x \in O \subset U$.

If instead of Hausdorffs axioms one takes the theorems 1 to 3 as axioms and theorem 4 as a definition for neighbourhood then one comes out with the usual definition of a topology:

Let $X$ be a set. A system $T$ of subsets of $X$ is called a topology on $X$, if:

1. $∅ \in T, X \in T$,
2. $O_1, O_2 \in T ⇒ O1 \cap O2 \in T$,
3. $S \subset T ⇒ (\bigcup_{S'∈S} S') ∈ T$.

This shows that "finite intersections aswell as arbitrary unions of open sets are open" is a generalization of the properties of "open sets" defined in terms of "neighbourhoods" that were defined to get a precise meaning of the idea of the convergence of a sequence.

Proof by example!

Consider the real numbers, and a sequence of ever-smaller open intervals around zero: (-1,1), (-1/2,1/2), (-1/3,1/3), ...

If we take any union of any of these sets, then any point in that union will have a little neighbourhood around it (can you see why?).

For intersections, it's not quite so simple.

If we take a finite number of the sets and take their intersection, then we'll still get an open set (hint: why is the intersection of two open sets still open?)

But if we take the infinite intersection, the only point that is in all the intervals is 0.

And 0 on its own isn't an open set, because 0 doesn't have a little neighbourhood around it.

If you understand why that's true, you should have no trouble seeing it more generally.

Perhaps this would be easier if it is motivated from the metric space context. If $(X,d)$ is a metric space, a subset $U \subset X$ is open if for all $x \in X$, there is some ball $B(x,r)$ of some radius $r > 0$, around $x$ such that $B(x,r) \subset U$. Now suppose we have a family of open sets $U_i$ for $i \in I$ some indexing set. The matter is fundamentally one of logical quantifiers.

(Unions) If we let $U$ be the union of the $U_i$, then if $x \in U$, we must have $x \in U_i$ for some single $i$. There is a ball $B(x,r) \subset U_i$. And by definition of a union $B(x,r) \subset U$. Crucial is that everywhere here we only need to verify "there exists some $i$ such that ..."

(Intersection) On the other hand $U$ is the intersection of $U_i$. If $x \in U$ then $x \in U_i$ for all $i$. We can find a ball $B(x,r_i) \subset U_i$ for each $i$. Let us assume this ball is chosen with maximal possible radius. If there was a ball $B(x,r) \subset U$ around $x$ then $B(x,r) \subset B(x,r_i)$ for each $i$ by the maximal radius assumption. Thus $r < r_i$,and $r \leq \inf_{i \in I}(r_i)$ would work. But now we have a problem because unless the index set $i$ is finite it is entirely possible the family of $\{r_i\}$ has $0$ as it's infimum. This is because to check $B(x,r)$ was contained in $U$ we needed to check $B(x,r) \subset U_i$ for all i.