Looking for a proof of an interesting identity

Using

$$\binom{n}{k}=\frac{1}{2 \pi i}\oint_C\frac{(1+z)^{n}}{z^{k+1}}dz$$ we get (integration contour is the unit cicrle)

$$ 2\pi iS_n=\oint dz \sum_{k=0}^{\infty}\frac{(1+z)^{n+2k}x^{n+2k}}{z^{k+1}2^{n+2k}}=\oint dz \frac{(1+z)^n x^n}{z2^n}\sum_{k=0}^{\infty}\frac{(1+z)^{2k}x^{2k}}{2^{2k}z^k}=\\ 4\frac{x^n}{2^n}\oint dz \underbrace{\frac{(1+z)^n}{4z-(1+z)^2x^2}}_{f(z)} $$

for $|x|<1$ only we have just one pole of $f(z)$ inside the unit circle namely $z_0(x)=\frac2{x^2}-\frac{2\sqrt{1-x^2}}{x^2}-1$ , so

$$ S_n=4\frac{x^n}{2^n}\text{res}(f(z),z=z_0(x))=4\frac{x^n}{2^n}\left[ \frac{1}{4 \sqrt{1-x^2}}\left(2\frac{1-\sqrt{1-x^2}}{ x^2}\right)^n\right] $$

or

$$ S_n=\frac{1}{\sqrt{1-x^2}}\left(\frac{1-\sqrt{1-x^2}}{ x}\right)^n $$

Extracting coefficients on the RHS we get the integral (coefficient on $x^{n+2k}$)

$$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+2k+1}} \frac{1}{\sqrt{1-z^2}} \left(\frac{1-\sqrt{1-z^2}}{z}\right)^n \; dz.$$

Now we put $(1-\sqrt{1-z^2})/z = w$ so that $z = 2w/(1+w^2).$ This has $w = \frac{1}{2} z + \cdots$ so the image in $w$ of the contour in $z$ can be deformed to a small circle enclosing the origin in the $w$-plane. (Moreover we see that the exponentiated term starts at $z^n$ which justifies the corresponding offset in the series.) We get $dz = 2/(1+w^2) - 4w^2/(1+w^2)^2 \; dw = 2(1-w^2)/(1+w^2)^2 \; dw.$ We also have $1-z^2 = 1 - 4w^2/(1+w^2)^2 = (1-w^2)^2/(1+w^2)^2.$ All of this yields

$$\frac{1}{2\pi i} \int_{|w|=\gamma} \frac{(1+w^2)^{n+2k+1}}{2^{n+2k+1} w^{n+2k+1}} \frac{1}{(1-w^2)/(1+w^2)} w^n \frac{2(1-w^2)}{(1+w^2)^2} \; dw \\ = \frac{1}{2^{n+2k}} \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{(1+w^2)^{n+2k}}{w^{2k+1}} \; dw.$$

This evaluates by inspection to

$$\frac{1}{2^{n+2k}} [w^{2k}] (1+w^2)^{n+2k} = \frac{1}{2^{n+2k}} [w^{k}] (1+w)^{n+2k} \\ = \frac{1}{2^{n+2k}} {n+2k\choose k}$$

which is the claim.