Problem: Let $ f: [ 0,1 ] \to \Bbb R $ be continuous function s.t. $ f(0) = f(1) $. Prove that for all natural $ n \geq 1 $ there exists $ x \in [ 0 ,1] $ s.t. $ f(x) = f(x + \frac{1}{n} ) $
Attempt: For $ n = 1 $ we'll choose $ x =0 $ and then we'll have $ f(0) = f(1) $. Let $ n\geq 2 $. If $ f $ is constant then we're finished. Assume $ f $ is not constant. Define $ g(x) = f(x) - f(x+\frac{1}{n}) $, $\forall x \in [0,1-\frac{1}{n}] $, note this function is continuous by the continuity of $ f $ on the given interval. [ I don't know how to proceed from here ]
Do you have any ideas as to how to proceed? I don't have any intuition as to what I should do next.
Thanks for the help in advance!