A simple graph must have at least one pair of vertices whose degrees are equal.

No, there is no assumption that the graph has a vertex of degree $0$: there is simply the observation that it is impossible for a graph on $n$ vertices to have both a vertex of degree $0$ and a vertex of degree $n-1$. If $G$ has a vertex $v$ of degree $n-1$, that vertex is adjacent to all of the other $n-1$ vertices of $G$, and therefore all of those vertices have degree at least $1$, because they all have an edge to $v$. Thus, $G$ cannot have both a vertex of degree $n-1$ and a vertex of degree $0$.

This means that if $G$ has a vertex of degree $n-1$, the only other possible degrees of its vertices are $1,2,\ldots,n-2$: degree $0$ is not possible. Thus, it can have at most the $n-1$ different degrees $1,\ldots,n-1$. And if it has a vertex of degree $0$, the only other possible degrees of its vertices are $1,2,\ldots,n-2$: degree $n-1$ is not possible. In this case it can have at most the $n-1$ degrees $0,\ldots,n-2$.

And of course if $G$ has neither a vertex of degree $0$ nor a vertex of degree $n-1$, the only possible degrees of its vertices are $1,\ldots,n-2$, so that it can have at most these $n-2$ degrees.

In every case, then, $G$ can have at most $n-1$ different degrees for its $n$ vertices, and the pigeonhole principle immediately tells us that two of the vertices must have the same degree.

There are $n$ vertices and there are $n$ possible degrees: $0,1,\ldots, n-1$. If no two vertices have same degree, then each possible degree must be attained. In particular, there must be one vertice $v$ of degree $0$ and one vertex $w$ of degree $n-1$. If $n>1$, this is a contradiction