How can I show that $\left|\sum_{n=1}^\infty\frac{x}{n^2+x^2}\right|\leq\frac{\pi}{2}$ for any $x\in{\bf R}$?

Consider the function $f(y) = \dfrac1{y^2+x^2}$. This is a monotonically decreasing function for $y>0$. Hence, we have $$\dfrac1{n^2+x^2} < \int_{n-1}^{n} \dfrac{dy}{y^2+x^2}$$ Hence, $$\sum_{n=1}^{\infty}\dfrac1{n^2+x^2} < \sum_{n=1}^{\infty}\int_{n-1}^{n} \dfrac{dy}{y^2+x^2} = \int_{0}^{\infty} \dfrac{dy}{y^2+x^2} = \dfrac{\pi}{2x}$$

You can rewrite the sum to $$ \frac {1}{x}\sum_{n=1}^\infty \frac{1}{(\frac n x)^2+1} $$ Because $f(n)=\frac1{n^2+1}$ is strictly decreasing for positive (reals) $n$, we know that \begin{align} \frac 1x\sum_{n=1}^\infty \frac{1}{(\frac nx)^2+1}&\leq \frac 1x\int_{n=0}^\infty\frac1{(\frac nx)^2+1} dn\\ &=\frac 1x \left(x \frac \pi 2\right)=\frac \pi 2 \end{align} For the integral, i used the substitution $m=\frac nx$, and because $dm=\frac 1x dn$, we get the factor $x$ in the result.

By showing that the sum is exactly

$$\frac{\pi}{2} \operatorname*{coth}{(\pi x)} - \frac1{2 x}$$

This may be done via, e.g., the residue theorem. Note that the expression is bounded from above by $\pi/2$ as $x \to \infty$