This is not a completely satisfactory answer. I would like a simpler one.
Nevertheless still probably a good exercise in Complex variables.
I will only sketch it.

What we want to show is equivalent to 
$$\zeta(2n)=-\frac{1}{2\pi i}\int_{C_r}\frac{\pi z \cot(\pi z)}{2z^{2n+1}}\,dz,\qquad n\ge 0,\quad n\in{\bf Z}.\eqno{(1)}$$
In fact this will be true for all $n\in{\bf Z}$. For $z=ix$ with $x>0$ we have
$$\cot(\pi z)=\cot(\pi i x)=-i-i\frac{2}{e^{2\pi x}-1}$$
it is convenient to write (1) as 
$$\zeta(2n)=-\frac{1}{2 i}\int_{C_r}\frac{ \cot(\pi z)+i}{2z^{2n}}\,dz,\qquad n\in{\bf Z}.\eqno{(2)}$$

Consider now the region $\Omega$ equal to ${\bf C}$ with a cut along the positive imaginary
axis. Let $\log z$ denote the determination of the logarithm in $\Omega$ with 
$-\frac{3\pi}{2}<\arg(z)<\frac{\pi}{2}$, and let $C'_r$ be the path of integration
that start at $i\infty$ to $ir$ (left border of the imaginary positive axis), then follows the circumference $C_r$ from $ir$ to 
$ir$ and then go from $ir$ to $i\infty$ (right border of the imaginary positive axis).
It is easy to show that (2) is equivalent to (3)
$$\zeta(2n)=-\frac{1}{2 i}\int_{C'_r}\frac{ \cot(\pi z)+i}{2z^{2n}}\,dz,\qquad  n\in{\bf Z}.
\eqno{(3)}$$

The integral defines  an entire function 
$$f(s)=-\frac{1}{4 i}\int_{C'_r}\bigl(\cot(\pi z)+i\bigr)e^{-s\log z}\,dz.\eqno{(4)}$$

When $\sigma=\Re(s)<0$ this can be transformed (let $r\to0$) in 
$$f(s)= e^{\pi i s/2}\sin(\pi s)(2\pi)^{s-1}\int_0^{\infty}\frac{x^{-s}}{e^{x}-1}\,dx,\qquad \sigma<0.\eqno{(5)}$$
Applying
Titchmarsh (2.4.1)
$$\zeta(s)=\frac{1}{\Gamma(s)}\int_0^\infty\frac{x^{s-1}}{e^x-1}\,dx,\qquad \sigma>1$$
and the functional equation, yields that for $\sigma<0$ we have
$$f(s)=e^{\pi i s/2}\cos(\pi s/2)\zeta(s)\eqno{(6)}$$
Therefore for all $s$ we have
$$e^{\pi i s/2}\cos(\pi s/2)\zeta(s)=-\frac{1}{4 i}\int_{\Gamma_r}\bigl(\cot(\pi z)+i\bigr)e^{-s\log z}\,dz.\eqno{(7)}$$
Since  $f(2n)=\zeta(2n)$ for all $n\in{\bf Z}$  we have proved that the coefficient
of $z^{2n}$ in the Laurent series for $-\frac{\pi z}{2}\cot(\pi z)$ is equal to 
$\zeta(2n)$ for all $n\in{\bf Z}$.