I'm stuck in proving the following: If the Ore's condition for a graph is fulfilled (i.e. $\text{deg(u)+deg(v)}\geq n$ for all non-adjacent vertices u,v), one can prove the following: For increasingly ordered sequences $(a_i)$ of the degrees in that graph G we have: $$\forall i<\frac{n}{2}: a_i\leq i \Rightarrow a_{n-i}\geq n-i$$.
My attempt was to start by contraction. That means there exist a_{i} and a_{n-i}, that does not fulfill this condition. Then according to the assumption, there is only one way, that this could happen, namely if there is an edge between x and y (corresponding to the degrees $a_i =$ d($x$) and $a_{n-i} =$ d($y$)). Then I started to take a look at one neighbouring vertex of x, which is not y and tried to do some case distinction, whether this neighbour is adjacent to y or not, but in the end I had nowhere to go ahead. Does anyone has a better idea than me? I would be really happy.
You are correct to start by assuming there is an $i$ for which the conclusion is false, and trying to get a contradiction.
Let $v_1, v_2, \dots, v_n$ be the vertices in order of increasing degree, so that $\deg(v_i) = a_i$.
Then if $a_i \le i < \frac n2$, then not just $v_i$ but $v_1, v_2, \dots, v_i$ must all have degree at most $i$. The sum of the degrees of any two of them is at most $2i < n$, so by Ore's condition, $v_1, v_2, \dots, v_n$ must all be adjacent.
This is the sort of reasoning you should employ. It's a starting point, but it doesn't get you all the way there; you will have to find some more edges that are forced to exist, and then convince yourself that there is a problem.