Proving that the intersection of any finite number of members of T belongs to T when (X,T) is a topological space. I know this is part of the definition of a topology. But the book I am using which is 'Topology without Tears' ask to prove this using mathematical induction.I hate to ask questions but proofs I have seen for this question in this site were different than mine.
Here it is...
The statement "Intersection of any two sets in $T$ belongs to $T$ " is given by the definition.
Suppose $T_1,T_2,T_3,...$ are members of $T$ and 
$$\bigcap_{n=1}^{k} T_n\in T \label{a}\tag{1}$$
is true.
 If $\,\bigcap_{n=1}^{k+1} \,T_n\in T $ is also true, the proposition can be proven by mathematical induction.
Since the element in the the left side of (\ref{a}) is a also set, let $M=\bigcap_{n=1}^{k} T_n $.
$$\bigcap_{n=1}^{k+1} T_n  = \bigcap_{n=1}^{k} T_n \cap T_{k+1} $$
$$\bigcap_{n=1}^{k+1} T_n = M \cap T_{k+1} $$
Since $M\in T $ by (1) and $T_{k+1}\in T $, $\,\bigcap_{n=1}^{k+1} T_n \in T$ by definition. Thus intersection of any finite number of sets in $T$ belongs to $T$. 
I want to know every mistake of this proof or ways to improve it.
 A: The proof by induction is fine. Renaming to $M$ is not really necessary of course (I don't usually introduce "unnecessary" notation), just writing 
$$\bigcap_{i=1}^{n+1} T_i = \left(\bigcap_{i=1}^{n} T_i\right) \cap T_{n+1}$$
(so using braces) is enough to make your point, I'd say.
It's actually a very common pattern: e.g. groups are closed under products of two members, but writing finite powers etc. is routinely done: having a set closed under an operation allows all finite "products' (or sums/intersections, etc. etc.) You do the proof once, and henceforth it's implicit.
It sometimes makes for slightly easier to read proofs, only having to check the $n=2$ case.
A minor nitpick: you don't mention the base case (which is $n=1$), which is almost too trivial to mention: $T_1$ in the topology gives us the $1$-ary intersection $\bigcap_{i=1}^1 T_i = T_1$ in the topology too, tautologically.  The $n=2$ case could also serve as the base case, depending on the formulation of the statement to be proved. But even the $0$-ary intersection (i.e. $X$) is in the topology.... 
