[Math] Intersection of arbitrary family of open sets is open: which step would be invalid in the proof of the finite case

Question

In other to show that

The intersection of a finite number of open sets is open.

We are doing the following:

Let $Q_1, …, Q_n$ be a open sets, and $x \in \bigcap_{i=1}^n Q_i$ be arbitrary. So by definition $\exists r_i>0$ s.t $B(x,r_i) \subset Q_i$, so define $r = min\{r1,…,r_n\}$, hence
$$B(x,r) \subset B(x, r_i) \subset Q_i \quad \forall i$$, hence $$B(x,r) \subset \in \bigcap_{i=1}^n Q_i,$$ and the intersection is open.

So my question is that if it was not a finite but an infinite intersection of subset, which step of the above proof would fail ?

Answer 1

Consider real numbers and $Q_i=(-1/i;1/i)$. But $$ \bigcap_i Q_i=\{0\} $$ and there is no space for a ball.