Finite intersection of open sets

Question

I would like to prove the following proposition:

Let $\tau$ be a topology. A finite intersection of elements of $\tau$ is also in $\tau$.

My attempt:

The proof is by induction on the number of elements in the intersection.

Base case: an element of $\tau$ is in $\tau$ by definition.

Suppose that the statement holds for intersections of $k\lt n$ elements. Let $S$ be an intersection of $n$ elements. The intersection of the first $n-1$ elements is in $\tau$ (by the induction hypothesis). Now we have an intersection of two elements of $\tau$ which is an element of $\tau$ (again by the induction hypothesis).

Is my proof correct?

Answer 1

The last step is not correct. The intersection of two open sets is open by definition of topology and not by induction hypothesis.

Consider this:

Theorem: If $A$ is a finite set, all of its elements are equal.

Proof: If $|A|=1$, the claim is trivially true. Suppose the statement holds for finite sets of $k<n$ elements. Let $A$ be a set of $n$ elements. Then the first $n-1$ elements are equal by induction hypothesis. Also the last two elements are equal by induction hypothesis. Therefore all $n$ elements are equal as was to be shown $_\square?$