A continuous periodic function is bounded

Idea. The theorem we need is the *extreme value theorem*: a continuous function on a closed bounded interval $[a,b]$ is bounded and attains its maximum and minimum. The function $f$ lives on all of $\mathbb{R}$, but periodicity lets us reduce to one period.

Reduction. On $[0,T]$, $f$ is continuous, so there exist $x_{\min},x_{\max}\in[0,T]$ with $f(x_{\min})\le f(t)\le f(x_{\max})$ for all $t\in[0,T]$.

Extension. Any $x\in\mathbb{R}$ can be written $x=t+kT$ with $t\in[0,T)$ and $k\in\mathbb{Z}$, and $f(x)=f(t)$. Therefore $f(x_{\min})\le f(x)\le f(x_{\max})$ for every real $x$.

Remark. Compare with $x\mapsto x\sin x$, which is continuous but not periodic and unbounded. Periodicity is exactly what allows the "compactness" of $[0,T]$ to spread over the whole line.