Let $(S,T)$ be a topological space. Verify that the intersection of any finite number of sets in $T$ is also in $T$ I am trying to learn some basic topology and came across this practice problem in the book I'm using. I'm not sure if my reasoning is correct:
Claim: For $n \geq 1$, we have the intersection of any $n$ sets in $T$ also belongs in $T$. 
Proof: By strong induction on $n$, the number of sets. 
BASE CASE: Let $n=1$. The intersection of a single set $S$ is simply $S$, so the claim trivially holds. 
INDUCTIVE STEP: We assume the claim holds for $n = 1,2,...,k$ and show the claim holds for $k+1$. 
Take any $k$ sets in $T$ and label them $S_1, S_2, ..., S_{k-1}, S_k$. Take another distinct set, call it $S_{k+1}$. We have the intersection of these $k+1$ sets can be written as follows:
$S_1 \cap S_2 \cap ... \cap S_{k-1} \cap S_{k} \cap S_{k+1} = (S_1 \cap S_2 \cap ... \cap S_{k-1})\cap(S_k \cap S_{k+1})$
By the inductive hypothesis for $n=k-1$ and $n=2$ on each of the sets above (respectively), we have the intersections are members of $T$. Call these intersectiosn $S_a, S_b$:
$=S_a \cap S_b$
By the inductive hypothesis with $n=2$, we have this intersection is in $T$ as well. QED
 A: This proof is incomplete.  The proof of the inductive step requires (in the last step) the case of $n=2$.  Since the $n=2$ case is one of the cases addressed in the inductive step, the inductive step is incomplete.  The case $n=2$ either must be explicitly added as a base step or the proof of the inductive step must not be made to rely on any of the $n > 1$ cases.
As a style issue : Many inductive proofs are actually "the first case has all the excitement and the subsequent cases are just doing the first case over and over again."  Your proof follows this structure, but you did not show all the excitement the first time it happens.  The $n=1$ case is vacuous, as you have shown.  The $n=2$ case is where the excitement happens and your proof shows this by the inductive step depending on the $n=2$ case.  Consequently, you should show your reader the $n=2$ case, so they can also see that all the excitement happens there.  Thus, there should be two base cases, the trivial $n=1$ case and the interesting $n=2$ case.  (I might not even describe the $n=1$ case as part of the induction.  I might dispose of it separately before even starting the induction.)
A: An arbitrary point in the intersection belongs to all of them.
For each open set you have an open ball centered at the point and contained in the set. 
This way you will find finitely many open balls centered at the point.
The ball with the minimum radius is contained in the intersection.
Thus the intersection is open.  
A: As you said in the comments, you're following the textbook "topology without tears" which has as one of its axioms:

The intersection of two sets from $\mathcal{T}$ is in $\mathcal{T}$.

Now you have to show by induction that any finite intersection is of sets from $\mathcal{T}$ is in $\mathcal{T}$ as well, so we show by induction
$$ \forall \mathcal{S} \subset \mathcal{T} : |\mathcal{S}| = n \implies \cap \mathcal{S} \in \mathcal{T}$$
As you said, the base statement for $n=1$ is trivial as then $\mathcal{S} = \{O\}$ with $O \in \mathcal{T}$ and then $\bigcap \mathcal{S} = O \in \mathcal{T}$.
If we assume the fact for $n$ and have $\mathcal{S}$ of size $n+1$ (for $n \ge 1$), we write $\mathcal{S} = \mathcal{S'} \cup \{O\}$ for some $\mathcal{S}$ of size $n$ and $O \in \mathcal{T}$. The induction hypothesis tells us that $$\bigcap \mathcal{S}' \in \mathcal{T}$$ and as
$$\bigcap \mathcal{S} = (\bigcap \mathcal{S}') \cap O$$
we see that $\bigcap \mathcal{S}$ is the intersection of two sets in $\mathcal{T}$, and by the topology axiom is in $\mathcal{T}$ as well. This proves the induction step.
