Consider the continuous and periodic function $f:\mathbb R \rightarrow \mathbb R$ with period $T > 0$ so that $f(x)=f(x+T)$ for any $x$.
Question: Prove that there exists a $c$ such that $f(c)=f(c+ \pi)$.
Note: This is not claiming that the period of $f$ is $\pi$, and is not a restatement of the definition of periodic. Many comments below reflect this incorrect interpretation of the statement of the problem.
Periodicity and continuity of $f$ implies boundedness, hence $f$ has a maximum and a minimum within the closure of a period $P=[0, T]$. Say the minimum and maximum are attained at $t_1\in P$ and $t_2\in P$. We have then
$$f(t_1) - f(t_1 + \pi) \leq 0 \quad f(t_2) - f(t_2 + \pi) \geq 0$$
as well as continuity of $f(t) - f(t + \pi)$.