All the roots of $\lambda x+\cot(x)=0$ are real. Looking for an alternative proof I am looking for an alternative (possibly simpler) proof of the following fact, that has some relevance in finding the eigenfunctions for the laplacian operator.

For any $\lambda\in\mathbb{R}^+$, $\lambda\geq 1$, all the solutions of 
  $$  \cot(x) + \lambda x = 0$$
  belong to $\mathbb{R}$.

My proof goes as follows: we just have to prove that all the roots of $f(x)=\cot(x)+\lambda x$ are real. If we assume the opposite, then $f'(x)=\lambda-\frac{1}{\sin^2(x)}$ must have some complex root by the Gauss-Lucas theorem. That is the same as stating that for some $z\in\mathbb{C}\setminus\mathbb{R}$ we have $\sin(z)=\pm\frac{1}{\sqrt{\lambda}}$, or, by setting $z=\sigma+it$,
$$ e^{i\sigma-t}-e^{-i\sigma+t} = \pm\frac{2i}{\sqrt{\lambda}}. $$
However, if we set $w=e^{iz}$, the equation $w-\frac{1}{w}=\frac{2i}{\sqrt{\lambda}}$ is solved only by 
$$ w = \frac{i\pm\sqrt{\lambda-1}}{\sqrt{\lambda}} $$
that is a number on the unit circle. That leads to $z\in\mathbb{R}$, i.e. $t=0$.

Extra question. What can we say about the distribution of the roots if $\lambda\in\mathbb{R}$ but $\lambda < 1$?

 A: Following the methodology outlined in the paper referenced by David Renfro, let $z=x+iy$ with $x\in \mathbb{R}$ and $y\in \mathbb{R}$.  Then, assume that $x\ne 0$ and $y\ne 0$.  The equation $\cot (z)+\lambda z=0$ implies that 
$$\frac{\sin(2x)}{\cosh(2y)-\cos(2x)}-i\frac{\sinh(2y)}{\cosh(2y)-\cos(2x)}+\lambda (x+iy)=0 \tag 1$$
Equating real and imaginary parts of $(1)$ reveals
$$\begin{align}
\frac{\sin(2x)}{\cosh(2y)-\cos(2x)}&=-\lambda x \tag 2\\\\\
\frac{\sinh(2y)}{\cosh(2y)-\cos(2x)}&=\lambda y \tag 3
\end{align}$$
Since neither $x$ nor $y$ is zero, the denominator of the left-hand sides of $(2)$ and $(3)$ are strictly positive.  We may divide, therefore, $(2)$ into $(3)$ to obtain
$$\frac{\sin(2x)}{\sinh(2y)}=-\frac{x}{y}$$ 
for $\lambda \ne 0$, whereupon rearranging yields
$$- 2 < \frac{\sin(2x)}{x}=-\frac{\sinh(2y)}{y} < -2\tag 4$$
We arrive at the desired contradiction.  Therefore, there can only be purely real or purely imaginary solutions to $\cot(z)+\lambda z=0$, $\lambda \ne 0$.
If $x=0$, then solutions of the equation of interest are solutions to 
$$\coth(y)=-\lambda y \tag 5$$
If we restrict $\lambda\ge 1$, then we see that there are no solutions to $\cot(iy)+\lambda (iy)=0$.

NOTE:
Special thanks to Martin R for identifying several errata in the original edit.
