Here is an approach:
We consider the following sum in the limit $n\rightarrow\infty$ (which we denote by $S$) :
$$
S_n=\frac{1}{\binom{2n}{n}}\sum_{k=0}^n(-1)^k\binom{2n}{n+k}\frac{1}{x^2+\pi^2 k^2}
$$
We split the range of summation at some $\delta$ such that $1<<\delta<<n$ to get $S_n=S_{2,n}+S_{1,n}$ and observe that $k<<n$ in $[0,\delta]$ so we can expand the binomial around small $k/n$ which means that
$$
S_{1,n}=\sum_{k=0}^{\delta}(-1)^k\left(1+O(k^2/n^3)\right)\frac{1}{x^2+\pi^2 k^2}
$$
since $|\sum_{k=0}^{\delta}(-1)^k\frac{k^2}{x^2+\pi^2 k^2}|<\sum_{k=0}^{\delta}\frac{k^2}{x^2+\pi^2 k^2}<\sum_{k=0}^{\delta}1=\delta$ we can bound the $O(k^2/n^3)$ term as follows
$$
S_{1,n}=\sum_{k=0}^{\delta}(-1)^k\frac{1}{x^2+\pi^2 k^2}+o(\delta/n^3)
$$
Taking limits we get that $S_{1,n}$ approaches a constant to determined later
$$
S_1=\sum_{k=0}^{\infty}(-1)^k\frac{1}{x^2+\pi^2 k^2}\,\,\,\, (*)
$$
Next we show that the tail sum $S_{2,n}$ approaches zero so that the limit in question is indeed given by $(*)$. To this end, observe that on $[\delta,n]$ we have $\pi k >> x$ so we can expand the fraction around large $k$. We find
$$
S_{2,n}=\frac{1}{\binom{2n}{n}}\sum_{k=\delta}^n(-1)^k\binom{2n}{n+k}\left(\frac{1}{\pi^2 k^2}+O(1/k^4)\right)
$$
since the binomial coefficient has a maximum at $k=0$ we can bound $\frac{\binom{2n}{n+k}}{\binom{2n}{n}}\leq1$ so
$$
|S_{2,n}|<\frac{1}{\pi^2}\sum_{k=\delta}^n\frac{1}{k^2}\sim \frac{1}{\pi^2} \int_{\delta}^n\frac{dk}{k^2}\sim \frac{1}{\pi^2\delta}
$$
so, choosing $\delta$ to be a (slowly enough) increasing sequence of $n$ we have indeed that
$$
S_2=0\,\,\,\,(**)
$$
Note that the $O(1/k^4)$ are vanishing even faster and can therefore also be neglected.
Putting anything $(*)$ and $(**)$ together we get
$$
S=S_1=\frac{1}{2x^2}+\frac{1}{2\sinh(x)x}
$$
were the last equality follows from Mittag-Lefflers Theorem (or the product expansion of sine)
Because $$\frac{1+(-1)^k}{2}=\begin{cases}1&\text{if $k$ is even}\\0&\text{if $k$ is odd}\end{cases}$$
we have $$\sum_{k\ge 0} a_{2k} = \sum_{k\ge 0} a_k \frac{1+(-1)^k}{2}.$$
Now take $a_k=\binom{n}{k+1}i^k$, where $i=\sqrt{-1}$, to obtain
\begin{align}
\sum_{k\ge 0} \binom{n}{2k+1}(-1)^k &= \sum_{k\ge 0} \binom{n}{2k+1}i^{2k} \\
&= \sum_{k\ge 0} \binom{n}{k+1}i^k \frac{1+(-1)^k}{2} \\
&= \frac{1}{2}\sum_{k\ge 0} \binom{n}{k+1}i^k + \frac{1}{2}\sum_{k\ge 0} \binom{n}{k+1} (-i)^k \\
&= \frac{1}{2i}\sum_{k\ge 0} \binom{n}{k+1}i^{k+1} - \frac{1}{2i}\sum_{k\ge 0} \binom{n}{k+1} (-i)^{k+1} \\
&= \frac{1}{2i}\sum_{k\ge 1} \binom{n}{k}i^k - \frac{1}{2i}\sum_{k\ge 1} \binom{n}{k} (-i)^k \\
&= \frac{1}{2i}\left((1+i)^n-1\right) - \frac{1}{2i}\left((1-i)^n-1\right) \\
&= \frac{(1+i)^n - (1-i)^n}{2i} \\
&= \frac{\left(\sqrt{2}(\cos(\pi/4)+i\sin(\pi/4)\right)^n - \left(\sqrt{2}(\cos(-\pi/4)+i\sin(-\pi/4)\right)^n}{2i} \\
&= \frac{\sqrt{2}^n(\cos(n\pi/4)+i\sin(n\pi/4)) - \sqrt{2}^n(\cos(-n\pi/4)+i\sin(-n\pi/4))}{2i} \\
&= \frac{\sqrt{2}^n(\cos(n\pi/4)+i\sin(n\pi/4)) - \sqrt{2}^n(\cos(n\pi/4)-i\sin(n\pi/4))}{2i} \\
&= \sqrt{2}^n \sin(n\pi/4)
\end{align}
Best Answer
We have $$(-1)^kk!(n-k)!=\dfrac{1}{n+2}\left[(-1)^k(k+1)!(n-k)!-(-1)^{k-1}k!(n+1-k)!\right]$$ Therefore \begin{align*}\sum_{k=0}^n\dfrac{(-1)^k}{n\choose k}&=\dfrac{1}{n!}\sum_{k=0}^n(-1)^kk!(n-k)!\\ &=\dfrac{1}{n!(n+2)}\left[(-1)^n(n+1)!0!-(-1)0!(n+1)!\right] \\ &=\dfrac{(n+1)![(-1)^n+1]}{n!(n+2)} \\ & = \dfrac{(n+1)[(-1)^n+1]}{n+2}\end{align*}