tan x = x on each branch

Monotonicity. On $I_k$, $\tan$ is differentiable and $\varphi'(x)=\frac1{\cos^2x}-1=\tan^2x\ge0$, with equality only at $x=k\pi$ (a single point). So $\varphi$ is strictly increasing on $I_k$.

Limits at the endpoints. As $x\to\left(k\pi-\frac\pi2\right)^+$, $\tan x\to-\infty$, so $\varphi\to-\infty$. As $x\to\left(k\pi+\frac\pi2\right)^-$, $\tan x\to+\infty$, so $\varphi\to+\infty$.

Conclusion. $\varphi$ is continuous and strictly increasing from $-\infty$ to $+\infty$ on $I_k$. By the intermediate value theorem it vanishes at least once, and by strict monotonicity at most once. So there is exactly one solution $x_k\in I_k$.

More precisely. Since $\varphi(k\pi)=-k\pi<0$, we have $x_k\in\left(k\pi,\,k\pi+\frac\pi2\right)$.