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I want to prove that at least two vertices have the same degree in any graph (with 2 or more vertices). I do have a few graphs in mind that prove this statement correct, but how would I go about proving it (or disproving it) for ALL graphs?

Say graph is simple with no loops and that all vertices have different degree $d_1,d_2,...d_n$, then $$d_1,d_2,...d_n = 0,1,2...,n-1$$

So there is a vertex with degree $n-1$ and a vertex with degree $0$. A contradiction.

I assume we're talking about finite graphs. I'm pretty sure your statement is false for infinite graphs.

Assume that a finite graph $G$ has $n$ vertices. Then each vertex has a degree between $n-1$ and $0$. But if any vertex has degree $0$, then no vertex can have degree $n-1$, so it's not possible for the degrees of the graph's vertices to include both $0$ and $n-1$. Thus, the $n$ vertices of the graph can only have $n-1$ different degrees, so by the pigeonhole principle at least two vertices must have the same degree.

EDIT: Graphs are typically defined as being finite. Infinite graphs are a generalization. I did not know this at the time of the post. I will leave this answer up though in case anyone finds it useful.

Here is a counterexample.

Let $G$ be a graph on the positive integers where there is an edge from $x$ to $y$ if $x < y le 2x$. Note that we will ignore the direction of the edges. So $2$, for example, is neighbored by $1$, $3$, and $4$.

Let $j$ and $k$ be distinct positive integers. Without loss of generality assume that $j < k$. Note that $deg(j) = j + lfloor j/2 rfloor$ and $deg(k) = k + lfloor k/2 rfloor$. We have that

$$j < k$$
$$lfloor j/2 rfloor le lfloor k/2 rfloor$$
$$j + lfloor j/2 rfloor < k + lfloor k/2 rfloor$$
$$deg(j) < deg(k)$$
$$deg(j) neq deg(k)$$

Therefore, no two different vertices will have the same degree. $square$