Proof that $\frac{(2n)!}{2^n}$ is integer

From 1 to $2n$ there are exactly $n$ even numbers. Hence the product $1\cdots 2n=(2n)!$ is divisible by $2\cdot2\cdots 2 \hbox{ ($n$ times)}= 2^n$.

Suppose there are $n$ distinct objects in a set $D$.

Consider a set $S$ containing $2$ copies of each element from $D$. Then $S$ has total $2n$ objects.

Total number of permuatations of these objects $=\dfrac{(2n)!}{(2!)^n}=\dfrac{(2n)!}{2^n}$.

Since number of permutations is an integer, therefore, $\dfrac{(2n)!}{2^n}$ is an integer.

Yes, you have proved it correctly. Indeed the proof is not difficult. If you need to do a formal induction, fine. But the result becomes obvious if you just expand, say, $(2\cdot 5)!$. It is clear that you pick up at least five $2$'s.

If you do need to write out a formal induction, it could be written out somewhat more clearly. For example, the phrase "so for $n=k$ it is true" is not clear. I assume you mean that "so if for $n=k$ it is true." We now write out a proof.

The result is obviously true for $n=1$. We show that if it is true for $n=k$, it is true for $n=k+1$.

Note that $$(2\cdot (k+1))!=(2k)!(2k+1)(2k+2).$$ By the induction assumption, $2^k$ divides $(2k)!$. It follows that $2^{k+1}$ divides $(2k)!(2)$, and therefore $2^{k+1}$ divides $(2k)!(2k+1)(2k+2)$.