It's quite hard to write down my solution, so it might be quite hard to understand. Suppose indirectly that we cannot choose a number for each vertex in the given way (*). We will replace $2012$ by $2n$. The vertices are: $A_1A_2\dots A_{2n}$. 1. Now choose randomly a number on vertex $A_1$. After that, choose a number for $A_2$ such that it isn't equal to the number on $A_1$, etc., continue this until $A_{2n}$. The only problem with this choice of numbers could be that the number chosen for $A_{2n}$ is equal to the number on $A_1$. 2. For convenience, denote the numbers written on vertex $A_i$ as $x_i$ and $y_i$. We already know that $x_1=x_{2n}$. Since this holds, we should switch $x_1$ to $y_1$. Then from (*) we must have, say, $x_2=y_1$. So switch $x_2$ to $y_2$ $\implies$ $y_2=x_3$. Etc. Finally, we end up with $x_{2n}=y_{2n-1}$, so we choose $y_{2n}$ for $A_{2n}$. This choice of numbers should be fine, the only problem could be that $y_{2n}=y_1$. 3. Since $y_1=y_{2n}$, rewrite $y_1$ to $x_1$. From (*), $x_1=y_2$; $A_2:y_2\to x_2$. (*) $\implies$ $x_2=y_3$, etc., $x_{2n-1}=y_{2n}$; $A_{2n}:y_{2n}\to x_{2n}$. Suppose that still, this doesn't work, or $x_{2n}=x_1$. Then $x_i=y_{i+1}$ from 3. and $y_i=x_{i+1}$ from 2. Hence, $x_{2n}=x_1=y_2=x_3=\dots=x_{2n-1}=y_{2n}$, which is a contradiction. Therefore, we have found a proper choice of numbers for the vertices (if we haven't already).

Alternatively, take two possible cases. Case 1: every vertex has the same pair of numbers. This case is trivial. Case 2: there are neighboring vertices with different pairs of numbers: A and B. Then choose a number on A different from those on B, and the rest is straightforward.