We notice that the vertices that occurs the same number of times in the sequence results in having the same degree. (Do I have to prove this?)
A short sentence about why this is true in general would be nice, yes.
Something like "since each appearance of a vertex in the sequence corresponds in one removal of an edge starting at that vertex, we see that the degree of the vertex is $(n-1)-k$, where $k$ is the number of appearances of the vertex in the sequence".
Anyway, your proof has a much bigger problem later on. In particular, you claim the following is true:
If some vertex occurs some j times less than (n-1) then by the nature of edges, at least one other edge must also occur j times.
yet you provide no proof. How do you know this is true? The way I see it, this claim is just an elaborate restatement of the original theorem that you are yet to prove, so you cannot just assert it is true.
Even more in particular, the problem is also here:
From this let's remove from the sequence one of the edges(so now it will occur n-2 times), in other words, we are removing one less edge. Then, one other edge also occurs n-2 times. So at least 2 vertices are of the same degree as the others essentially occur the same time.
well... sure. But all this shows is that if you remove one edge from a full graph, you get two vertices with degree $n-2$.
But then you say:
We could do this repeatedly
How do you know this? You demonstrated the claim on a very specific type of graph (i.e., a full graph), and then claim that the same is true on all graphs. You cannot do that.
Edit:
You now also demonstrated the claim on another specific type of graphs (those that can be constructed from a full graph by removing one edge), but you still haven't shown it on all graphs.
Edit #2
Your new approach is a good idea, but it needs some work. The way you did it is ok, but can be made a little clearer. When analyzing the case when $k$ vertices have degree $0$, you can first say that the only issue appears when $k=1$ (otherwise, two vertices have degree $0$ and we are done), and then explain that if one vertex has degree $0$, then the maximum degree of the others is $n-2$.
Best Answer
Suppose such a graph exists. Each vertex in the graph can have a degree from $0$ to $n-1$ (simple graphs do not forbid a degree-$0$ vertex, connected graphs do). Since this range spans $n$ values in total and each vertex degree is different, the degrees are distributed one-per-vertex. In particular, there must exist a vertex with degree $n-1$ and one with degree $0$.
Now the degree-$n-1$ vertex is connected to all other vertices because the graph is simple, including the degree-$0$ vertex. But the latter is not connected to any vertex, which is a contradiction. Therefore two vertices in the graph must have the same degree.