Show that $\sum\limits_{k=0}^n\binom{2n}{2k}^{\!2}-\sum\limits_{k=0}^{n-1}\binom{2n}{2k+1}^{\!2}=(-1)^n\binom{2n}{n}$

Idea. We shall obtain the identity by equating the coefficient of $x^{2n}$ in the expansions of $(1+x^2)^{2n}$ and $(1+ix)^{2n}(1-ix)^{2n}$.

The binomial expansion of $(1+x^2)^{2n}$ is $$ (1+x^2)^{2n}=\sum_{k=0}^{2n}\binom{2n}{k}x^{2k}, $$ while $$ (1+x^2)^{2n}=(1+ix)^{2n}(1-ix)^{2n}=\left(\sum_{k=0}^{2n}\binom{2n}{k}(ix)^{k}\right)\left(\sum_{k=0}^{2n}\binom{2n}{k}(-ix)^{k}\right). $$ The coefficient of $x^{2n}$ in the first expansion is $a_{2n}=\displaystyle\binom{2n}{n}$, while in the right-hand side of the above it can be expressed as the following sum: \begin{align} a_{2n}&=\sum_{k=0}^{2n}i^k(-i)^{2n-k}\binom{2n}{k}\binom{2n}{2n-k}=i^{2n} \sum_{k=0}^{2n}(-1)^{k}\binom{2n}{k}\binom{2n}{k} \\&=(-1)^{n} \sum_{k=0}^{2n}(-1)^{k}\binom{2n}{k}^{\!2}= (-1)^{n}\sum_{k=0}^{2n}(-1)^k\binom{2n}{k}^{\!2}\\ &= (-1)^{n}\sum_{k=0}^{n}\binom{2n}{2k}^{\!2}-(-1)^{n}\sum_{k=0}^{n-1}\binom{2n}{2k+1}^{\!2}\!\!. \end{align} Hence $$ \sum_{k=0}^n\binom{2n}{2k}^2-\sum_{k=0}^{n-1}\binom{2n}{2k+1}^2=(-1)^n\binom{2n}{n}. $$

This is simpler if you rewrite your left hand side as $$ \sum_{k=0}^n\binom{2n}{2k}^2-\sum_{k=0}^{n-1}\binom{2n}{2k+1}^2= \sum_{i=0}^{2n}(-1)^i\binom{2n}i^2= \sum_{i=0}^{2n}(-1)^i\binom{2n}i\binom{2n}{2n-i} $$ first. So you are looking for the coefficient of $X^{2n}$ in the product $(1-X)^{2n}(1+X)^{2n}=(1-X^2)^{2n}$, which is the same as the coefficient of $Y^n$ in $(1-Y)^{2n}$. That coefficient is clearly $(-1)^n\binom{2n}n$.

Note that one could replace $2n$ by $m$ and allow it to be odd, if one takes the right hand side to be$~0$ in that case. The proof remains the same, remarking that there is obviously no term $X^m$ in $(1-X^2)^m$ when $m$ is odd.

Let me also give a combinatorial proof (since that tag is given). Given $m$ mixed-sex couples, the left hand side counts the number of gender-balanced subsets (containing equally many women as men), counted with a factor$~{-}1$ if that number for each sex is odd. In this counting, the subsets that do not contain exactly one member of each couple cancel out as follows: for such a subset, find the first couple of which which not exactly one member was selected, and add or remove the couple to obtain a cancelling subset; this operation is clearly a gender parity respecting involution. So we are left with as un-cancelled contributions those from the gender-balanced subsets with one member from each couple, which subsets are clearly of size$~m$. If $m$ is odd then no subsets at all remain, as gender balance is impossible. If $m$ is even then each remaining gender-balanced subset is counted with sign $(-1)^{m/2}$and it is determined by the $m/2$ women it contains (it is completed by the men from the other $m/2~$couples), so there are $\binom m{m/2}$ of them.