Universal Chord Theorem Let $f \in C[0,1]$ and $f(0)=f(1)$. 
How do we prove $\exists a \in [0,1/2]$ such that $f(a)=f(a+1/2)$?
In fact, for every positive integer $n$, there is some $a$, such that $f(a) = f(a+\frac{1}{n})$.
For any other non-zero real $r$ (i.e not of the form $\frac{1}{n}$), there is a continuous function $f \in C[0,1]$, such that $f(0) = f(1)$ and $f(a) \neq f(a+r)$ for any $a$.
This is called the Universal Chord Theorem and is due to Paul Levy.
Note: the accepted answer answers only the first question, so please read the other answers too, and also this answer by Arturo to a different question: https://math.stackexchange.com/a/113471/1102

This is being repurposed in an effort to cut down on duplicates, see here: Coping with abstract duplicate questions.
and here: List of abstract duplicates.
 A: I have first encountered this result in the book Van Rooij, Schikhof: A Second Course on Real Functions.
I will copy here the text of Exercise 9.P.

Let $0<\alpha<1$. Let $f\colon[0,1]\to\mathbb R$, $f(0)=f(1)$.
  (i) Show that if $\alpha$ is one of the numbers $\frac12,\frac13,\frac14,\dots$ and if $f$ is continuous, then the graph of $f$ has a horizontal chord of length $\alpha$, i.e., there exists $s,t\in[0,1]$ with $f(s)=f(t)$ and $|s-t|=\alpha$. 
  (ii) The proof you gave probably relies on Darboux continuity. Prove, however, that the given continuity condition on $f$ may not be weakened to Darboux continuity. (Take $\alpha:=\frac12$ and start with a function on $(0,\frac12]$ that maps every subinterval of $(0,\frac12]$ onto $\mathbb R$.)
  (iii) Now let $\alpha\notin\{\frac12,\frac13,\dots\}$. Define a continuous function on $[0,1]$ with $f(0)=f(1)$ whose graph has no horizontal chord of length $\alpha$. (Choose $f$ such that $f(x+\alpha)=f(x)+1$ for $x\in[0,1-\alpha]$.) 

For this questions only the first and the third part are relevant. And the first part has been already solved in other answers.
Let us spell out in details construction following the hint from the third part.  (Although the hint given there already gives quite a good idea how to proceed.) This is slightly different from the examples given in other answers.
We want to have $f(x+\alpha)=f(x)+1$. Notice that this also implies $f(x+k\alpha)=f(x)+k$.
If we define the function on the interval $[0,\alpha]$, then the above condition determines the function $f$ uniquely on the rest of the interval $[0,1]$.
We want to have $f(0)=0$ and $f(\alpha)=f(1)$.
Let us denote $n=\left\lfloor\frac1\alpha\right\rfloor$, i.e., $n$ is the largest integer such that $n\alpha<1$.
Then we have $1-n\alpha\in(0,\alpha)$. We need to choose $f(1-n\alpha)=-n$ in order to get $f(1)=0$.
(Notice that this cannot be done if $n\alpha=1$. It is possible only if $0<1-n\alpha<\alpha$, since the values $f(0)$ and $f(1)$ are already prescribed.)
Then arbitrary function defined as above (i.e., with the prescribed values in the points $0$, $1-n\alpha$, $\alpha$ and extended from $[0,\alpha]$ to the whole interval using $f(x+\alpha)=f(x)+1$) satisfies the required conditions. 
Such function for a specific choice of $\alpha$ is illustrated in this picture:

For comparison, here is plot of Lévy's function mentioned in Aryabhata's answer for the same value of $\alpha$ can be checked on WolframAlpha.
A: Hint:   consider $g(x)$, defined on $[0,1/2]$ by $g(x)=f(x+1/2)-f(x)$
A: Interestingly,
The numbers of the form $r = \displaystyle \frac{1}{n} \ \ n \ge 1$ are the only positve numbers such that for any continuous function $\displaystyle f:[0,1] \to \mathbb{R}$ such that $\displaystyle f(0) = f(1)$, there is some point $\displaystyle c \in [0,1-r]$ such that $\displaystyle f(c) = f(c+r)$. 
For any other positive $r$ we can find such a continuous function for which there is no $c$ such that $f(c) = f(c+r)$.
For a proof that $\displaystyle r = \frac{1}{n}$ satisifies this property, let $\displaystyle g(x) = f(x) - f(x+ \frac{1}{n})$, for $\displaystyle x \in [0, 1-\frac{1}{n}].$
Then we have that $\displaystyle  \sum_{k=0}^{n-1} \ g\left(\frac{k}{n}\right) = 0$.
Thus, if none of $\displaystyle g\left(\frac{k}{n}\right)$ are $\displaystyle 0$, then  $\displaystyle \exists i,j \in [0, 1, ..., n -1] \ni \displaystyle g\left(\frac{i}{n}\right) \gt 0$ and $\displaystyle g\left(\frac{j}{n}\right) \lt 0$.
For any positive $\displaystyle r$, consider the following example, due to Paul Levy.
$\displaystyle f(x) = \sin^2\left(\frac{\pi x}{r}\right) - x \ \sin^2\left(\frac{\pi}{r}\right)$. Clearly, $f$ is continuous and $f(0)=0=f(1).$
If $\displaystyle f(x) = f(x+r)$, then, $\displaystyle r\ \sin^2\left(\frac{\pi}{r}\right) = 0$ and hence, $\displaystyle r = \frac{1}{m}$ for some integer $\displaystyle m$.
Apparently this is called the Universal Chord Theorem (due to Paul Levy!).
A: You want to use the intermediate value theorem, but not applied to $f$ directly. Rather, let $g(x)=f(x)-f(x+1/2)$ for $x\in[0,1/2]$. You want to show that $g(a)=0$ for some $a$. But $g(0)=f(0)-f(1/2)=f(1)-f(1/2)=-(f(1/2)-f(1))=-g(1/2)$. This gives us the result: $g$ is continuous and changes sign, so it must have a zero.
