Proving that $\sum_{k=0}^{2n} {2k \choose k } {2n \choose k}\left( \frac{-1}{2} \right)^k=4^{-n}~{2n \choose n}.$ I have happened to have  proved this sum while attempting to prove another summation. Let
$$S_n=\sum_{k=0}^{2n}  {2k \choose k }  {2n \choose k}\left( \frac{-1}{2} \right)^k$$
${2k \choose k} $ is the coefficient of $x^0$ in $(x+1/x)^{2k}$.
Consequently, $S_n$ is the coefficient of $x^0$ in $$f(x)= \sum_{k=0}^{2n} {2n \choose k}~\left (x+\frac{1}{x}\right)^{2k}~\left ( \frac{-1}{2}\right)^{k}=\left(1-\left(\frac{x+1/x}{\sqrt{2}}\right)^2 \right)^{2n} = 4^{-n} ~ \left(x^2+\frac{1}{x^2}\right)^{2n}.$$ Finally, the coefficient of $x^0$ 
in $f(x)$ is $$4^{-n}~{2n \choose n}=S_n.$$
I hope that you will find it interesting and prove it in some other way. Do try!

We use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series. Recalling the generating function of the central binomial coefficients we can write 
\begin{align*}
[z^n]\frac{1}{\sqrt{1-4z}}=\binom{2n}{n}\tag{1}
\end{align*}

We obtain
  \begin{align*}
\color{blue}{\sum_{k=0}^{2n}}&\color{blue}{\binom{2n}{k}\binom{2k}{k}\left(-\frac{1}{2}\right)^k}\\
&=\sum_{k=0}^{2n}\binom{2n}{k}[z^k]\frac{1}{\sqrt{1+2z}}\tag{2}\\
&=[z^0]\frac{1}{\sqrt{1+2z}}\sum_{k=0}^{2n}\binom{2n}{k}z^{-k}\tag{3}\\
&=[z^0]\frac{1}{\sqrt{1+2z}}\left(1+\frac{1}{z}\right)^{2n}\tag{4}\\
&=[z^{-1}]\frac{(1+z)^{2n}}{z^{2n+1}\sqrt{1+2z}}\tag{5}\\
&=[t^{-1}]\frac{\left(1+\frac{t}{1-t}\right)^{2n}}{\left(\frac{t}{1-t}\right)^{2n+1}\sqrt{1+\frac{2t}{1-t}}}\cdot\frac{1}{(1-t)^2}\tag{6}\\
&=[t^{2n}]\frac{1}{\sqrt{1-t^2}}\tag{7}\\
&\,\,\color{blue}{=\frac{1}{4^n}\binom{2n}{n}}\tag{8}
\end{align*}

Comment:


*

*In (2) we apply the coefficient of operator according to (1).

*In (3) we use the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.

*In (4) we apply the binomial theorem.

*In (5) we write the expression using formal residual by applying again the rule from comment (3).

*In (6) we  use the substitution  $z=\frac{t}{1-t}, dz=\frac{1}{(1-t)^2}dt$.

*In (7) we do some simplifications.

*In (8) we select the coefficient of $t^{2n}$ by taking  (1)  evaluated  at  $z=\frac{1}{4}t^2$.
