For $n\ge4$ If every vertex has an even degree, then at least 3 vertices have the same degree. Let $G$ be a simple undirected graph (no loops nor multi edges), $n=|v|$
My try:
Let $A,B,C,D$ be my vertices with no edges between them, their $deg.=0$ so the condition holds.
I can assign only $\large\lceil\frac{n}{2}\rceil$ distinct $degrees$ because every $v$ has an $even\space degree$.
But how to show that at least $3 \space vertices $ have same $degree ?$ (I can draw it by adding edge after edge to my nodes but it doesn't give me any new clues)
 A: If there are $2n$ vertices, then the possible degrees are $2, 4, \ldots 2n - 2$, i.e. we have $n - 1$ unique possible degrees. We have to spread $n-1$ unique degrees among $2n$, so it is clear that at least 3 vertices should have same degree by pigeonhole principle.
If there are $2n + 1$ vertices, then the possible degrees are $2, 4, \ldots, 2n$, i.e. $n$ unique possible degrees. WE have to spread $n$ unique degrees among $2n + 1$ verticies, so it is clear even in this case that at least 3 vertices should have same degree by pigeonhole principle.
EDIT: The OP asked why I am not considering vertices of degree 0. This was an oversight. It can be fixed in this way. If there are three or more vertices of degree 0, we are done. By the above argument we are also done when there are no vertices of degree 0.
So all we have left are the cases of 1 or 2 vertices of degree 0.
Case 1 (two vertices of degree 0):
Case 1a) 2n vertices. Degree of 2n-2 is not possible because there are two vertices of degree 0. Same pigeonhole principle applies.
Case 1b) 2n+1 vertices. Degree of 2n is not possible because there are two 
vertices of degree 0. Sample pigeonhole principle applies.
Case 2 (one vertex of degree 0):
Case 2a) 2n vertices. One vertex has degree 0. So we have 2n-1 vertices left and degrees of 2, 4, ... 2n-2 to spread. Thus we have n-1 unique degrees to spread amongst 2n-1 vertices. pigeonhole principle...
Case 2b) 2n+1 vertices. One vertex has degree 0. So we have 2n vertices left and degrees of 2, 4, ... 2n-2 to spread. pigeonhole principle again...
A: If you draw all the graphs with 4 vertices such every vertex has degree even then you see it is true. Now with induction suppose the problem is true for all graphs with at most $K$ vertices. Suppose $G$ is a graph with $K+1$ vertices such every vertex has degree even. If a vertex in $G$( for example $v$) has degree $0$ then suppose the graph $G-v$ and with induction it has three vertices of same degree. Suppose no vertex of $G$ has degree zero. We have two condition:
If the number of vertices of $G$ is $2p+1$ then the degrees come from ${2,\cdots ,2p}$ now with Pigeonhole principle at least three of them have same degree. 
If the number of vertices of $G$ is $2p$ then the degrees come from ${2,\cdots ,2p-2}$ and again use Pigeonhole principle.
