Note that $\arctan(\cdot)$ is increasing on $[0,1]$ and for $x \geqslant 1$ and $y > 0$ we have
$$0 \leqslant \frac{2y}{x^2 + y^2} \leqslant \frac{2xy}{x^2 + y^2} \leqslant 1$$
Thus,
$$\sup_{y \in \mathbb{R^+}}\int_n^\infty \arctan\frac{2y}{x^2 + y^2} \, dx > \sup_{y \in \mathbb{R^+}}\int_n^{2n} \arctan\frac{2y}{x^2 + y^2} \, dx \\\ > \sup_{y \in \mathbb{R^+}}\left(n\arctan\frac{2y}{(2n)^2 + y^2}\right) \\\ \geqslant n\arctan\frac{2n}{4n^2 + n^2 } \\\= n\arctan\frac{2}{5n } $$
Using the Taylor expansion $\arctan x = x + \mathcal{O}(x^3)$ for $x \in (-1,1]$ we have
$$\sup_{y \in \mathbb{R^+}}\int_n^\infty \arctan\frac{2y}{x^2 + y^2} \, dx > n\left[\frac{2}{5n} + \mathcal{O}\left(\frac{1}{n^3} \right)\right] \\\ = \frac{2}{5} + \mathcal{O} \left(\frac{1}{n^2} \right)$$
Since the RHS does not converge to $0$ as $n \to \infty$, the convergence of the improper integral is not uniform for $y \in [0,\infty)$ and, consequently, for $y \in \mathbb{R}$.