We need to divide $2n$ members of the club into $n$ disjoint teams of $2$ members each means, $i \ne j \Longrightarrow \text{Team}_i \cap \text{Team}_j = \phi$ and $|\text{Team}_k| = 2$.
Let $A = \{a_1,a_2, \dots, a_{2n}\}$ be the members of the club.
Let $\pi(A)$ be the permutation of $A$, such that total possible permutation be $(2n)!$. Each permutation can be represented as $\pi_i(A)$ for $1 \le i \le 2n$.
For example, $n=3$
One such $\pi_k(A) = a_3a_6a_1a_4a_5a_2$, then we select first two $a_i$'s and make a team of them, then we select next two $a_i$'s and make a team of them and so on.
$$\begin{array}{|c|c|c|}
\hline
a_3a_6 & a_1a_4 & a_5a_2\\
\hline
\text{Team}_1 & \text{Team}_2 & \text{Team}_3 \\
\hline
\end{array}$$
Now $\pi_j(A) = a_6a_3a_1a_4a_5a_2$,
$$\begin{array}{|c|c|c|}
\hline
a_6a_3 & a_1a_4 & a_5a_2\\
\hline
\text{Team}_1 & \text{Team}_2 & \text{Team}_3 \\
\hline
\end{array}$$
Though $\pi_k(A) \neq \pi_j(A)$, yet the teams formed using $\pi_k(A)$ and $\pi_j(A)$ are same!! Since, order of members in the team doesn't matter. Hence, we should divide by $2$ for every $\text{Team}_i$.
In total there are going to be $(2n)!$ such sequences of $2n$ members and $n$ such teams made on every such sequence, and for every such team we are dividing the total ways by $2$.
$$(2n)!\frac{1}{2^n}$$
We are also given one more condition to satisfy, i.e. "The teams are not labelled". Until now, we have a working solution for labelled teams, so now lets take $\pi_p(A), \pi_q(A)$ for the above mentioned example to understand this condition.
$\pi_p(A) = a_1a_4a_3a_6a_5a_2$
$\pi_q(A) = a_5a_2a_1a_4a_3a_6$
$$\begin{array}{|c|c|c|c|}
\hline
\pi_k(A) & a_3a_6 & a_1a_4 & a_5a_2\\
\hline
\pi_p(A) & a_1a_4 & a_3a_6 & a_5a_2\\
\hline
\pi_q(A) & a_5a_2 & a_1a_4 & a_3a_6\\
\hline
\vdots & a_5a_2 & a_3a_6 & a_1a_4\\
\hline
\vdots & a_1a_4 & a_5a_2 & a_3a_6\\
\hline
\vdots & a_3a_6 & a_5a_2 & a_1a_4\\
\hline
& \text{Team}_1 & \text{Team}_2 & \text{Team}_3 \\
\hline
\end{array}$$
This shows that members partition into team matters and not the team order itself. Above $6$ ways are effectively $1$ way only, hence we need to divide total ways with $n!$ since there are $n$ teams.
Final answer would be
$$\dfrac{(2n)!}{2^n (n)!}$$
$\textbf{Option (C) is correct}$.