Let $f: \mathbb{R} \to \mathbb{R}$ be continuous periodic function with period $T>0$

The simplest approach is: Consider $g(x) = f(T/2+x) - f(x).$ Now if $g(0)$ is positive, then $g(T/2)$ has to be negative (since $g(x) + g(x+T/2) = f(T+x) - f(x) =0.$) The result then follows by the intermediate value theorem (as you have guessed).

I want to answer the more general case posed above. Let $\alpha \in \mathbb{R}$. We want to show that there exists a $y\in\mathbb{R}$ such that $f(y+\alpha)=f(y)$. Sometimes in order to show two functions are equal, it helps to consider a function that is the two functions subtracted.

Consider $g(y)=f(y+\alpha)-f(y)$. We know that periodic continuous functions have a max and min value. Let $x_0$ be the max. Then,

1) $g(x_0)=f(x_0+\alpha)-f(x_0)\leq 0$ and

2) $g(x_0-\alpha)=f(x_0)-f(x_0-\alpha)\geq 0$.

(1) and (2) are true since $f(x_0)$ is the max of the function. Now if either (1) or (2) equal zero, we are done and have found our value $y$ such that $f(y+\alpha)=f(y)$. If neither of them are zero, then we can use the Intermediate Value Theorem to show that there exists a $y\in [x_0-\alpha,x_0]$ such that $g(y)=0$. This implies that $f(y+\alpha)=f(y)$.