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 will use
$$
\frac1{1-x}\left(\frac{x}{1-x}\right)^k=\sum_{n=0}^\infty\binom{n}{k}x^n\tag1
$$
and
$$
(1-4x)^{-1/2}=\sum_{k=0}^\infty\binom{2k}{k}x^k\tag2
$$
Extracting the even part of $(1)$
$$
\begin{align}
\sum_{n=0}^\infty\binom{2n}{k}x^{2n}
=\frac12\left[\frac1{1-x}\left(\frac{x}{1-x}\right)^k+\frac1{1+x}\left(-\frac{x}{1+x}\right)^k\right]\tag3
\end{align}
$$
Compute the generating function of the sum we want
$$
\begin{align}
&\sum_{n=0}^\infty\sum_{k=0}^\infty\binom{2k}{k}\binom{2n}{k}\left(-\frac12\right)^kx^{2n}\tag4\\
&=\frac1{2-2x}\sum_{k=0}^\infty\binom{2k}{k}\left(-\frac{x}{2-2x}\right)^k
+\frac1{2+2x}\sum_{k=0}^\infty\binom{2k}{k}\left(\frac{x}{2+2x}\right)^k\tag5\\
&=\frac1{2-2x}\left(1+\frac{4x}{2-2x}\right)^{-1/2}
+\frac1{2+2x}\left(1-\frac{4x}{2+2x}\right)^{-1/2}\tag6\\
&=\frac1{2-2x}\left(\frac{1-x}{1+x}\right)^{1/2}
+\frac1{2+2x}\left(\frac{1+x}{1-x}\right)^{1/2}\tag7\\[6pt]
&=\left(1-x^2\right)^{-1/2}\tag8\\[6pt]
&=\sum_{n=0}^\infty\frac1{4^n}\binom{2n}{n}x^{2n}\tag9
\end{align}
$$
Explanation:
$(4)$: compute the generating function
$(5)$: apply $(3)$
$(6)$: apply $(2)$
$(7)$: simplify
$(8)$: simplify
$(9)$: apply $(2)$ 
Equating coefficients of $x^{2n}$ gives
$$
\sum_{k=0}^\infty\binom{2k}{k}\binom{2n}{k}\left(-\frac12\right)^k
=\frac1{4^n}\binom{2n}{n}\tag{10}
$$
