Proof that the intersection of any finite number of elements of $\tau$ is a member of $\tau$, if $(X,\tau)$ is a topological space. Let $(X,\tau)$ be any topological space, verify that the intersection of any finite number of elements of $\tau$ is a member of $\tau$. Prove by mathematical induction.
My trial is attached this post. What is missing in my proof?
I feel it's not balanced yet.
Proof:  We are given that $(X, \tau)$ is a topological space and we are to show that the intersection of any finite number of elements of $\tau$ is a member of $\tau$.
Using mathematical induction, let $x_n \in X$ and $ \{ x_n \} \in \tau$.  Let $n=1$, so that $x_1 \in X$ and $ \{ x_1 \} \in \tau$ is true.  Let $n=k$, so that $x_k \in X$ and $ \{ x_k \} \in \tau$ is true.  And let $n=k+1$ so that $x_{k+1} \in X$ and $ \{ x_{k+1} \} \in \tau$ us also true.  Since $ \{ x_{k+1 \} \in \tau$, let
$$A = \cap_{k=1}^{\infty} x_{k+1}.$$
Since each single set belongs to $\tau$, then the intersection of every singleton will also belong to $\tau$; i.e., $A \in \tau$ and $$A= \{ x_1 \} \cap \{ x_2 \} \cap \{ x_3 \} \cap \ldots \cap_k^{\infty} \{ x_{k+1} \}.$$

 A: Your proof is off in a couple of respects.  First, any open set is a potential member of the intersection.  Not only is it possible that there are open sets that aren't singletons, in fact in most topologies that you run across, singletons won't be open sets.  Second, it's simply not true that an infinite intersection of open sets (i.e., sets in the topology) is necessarily open (i.e., in the topology).
We often designate open sets in a topology by $U$ and $V$.  Assume $U_1, U_2 \in \tau$.  Then by the definition of a topology, you know that $U_1 \cap U_2 \in \tau$.  That's the base case of your induction.
Now assume $U_1, U_2, U_3 \in \tau$.  Then $$U_1 \cap U_2 \cap U_3 = (U_1 \cap U_2) \cap U_3.$$  We know from the previous step that $U_1 \cap U_2 \in \tau$ and we've assumed $U_3 \in \tau$, so the right-hand side is an intersection of two sets that are in $\tau$, so it also must be in $\tau$.  Thus, $U_1 \cap U_2 \cap U_3 \in \tau$.
Do you see how to complete the inductive step now?  If not, try to mimic the proof for $k=4$ and you should get there.  As a bonus, do you see why this proof doesn't work for infinite intersections?
A: A model solution following Robert's hint:
We'll prove by induction on $n \ge 1$ that
$$\forall O_1, \ldots O_n \in \tau: \bigcap_{i=1}^n O_i \in \tau\tag{$\phi_n$}$$
holds.
For $n=1$ the statement is trivial. $O_1 \in \tau$ implies $O_1 \in \tau$.
Suppose $\phi_n$ holds for some $n \ge 1$. To see that $\phi_{n+1}$ then also holds, let $O_1, \ldots, O_{n+1} \in \tau$ be arbitrary. Then
$$\bigcap_{i=1}^{n+1} O_i = \left( \bigcap_{i=1}^n O_i\right) \cap O_{n+1}\tag{1}$$
By $\phi_n$ we know that $\bigcap_{i=1}^n O_i \in \tau$ and so by the intersection axiom for topologies it follows that $\bigcap_{i=1}^{n+1} O_i \in \tau$ too, so $\phi_{n+1}$ also holds.
So we've proved the statement by induction on $n$.
If we're really going wild, I could run the induction from $n=0$ onwards, using the common convention that $\bigcap_{i=1}^0 O_i = X$ (intersection of empty family) and it still would work out fine as $X \in \tau$ for any topology and $(1)$ also holds for $n=0$ in that case..
