Please see edit.
I am currently stuck at a proof where to my understanding the key step is to argue how the fact that if in a subgraph, the degree of each vertex is at least the size of the largest clique in the graph $G$, $d(v) \geq \omega(G), v \in V(G)$, then the original assumption of ... is false. My guesses are that
1.) the contradiction is something like: "if $d(v) \geq \omega(G)$, then the subgraph contains a clique larger than the largest clique in the original graph".
2.) contradiction can be constructed somehow with pigeonhole principle from the fact that i.) the vertex set $V(G)$ is finite, ii.) self-loops are not allowed, iii.) each vertex $v$ is connected to $\omega(G)$ other vertices.
My current problem is that I don't know how to argue about the contradiction in general. I can give a constructive argument about a single graph, in which the aforementioned facts lead to a contradiction, but I don't know whether this is enough to argue in general.
The constructive argument about a single graph is that suppose there are in total $\omega(G) + 2$ nodes. Partition the nodes as $T_1 = \{r\}, T_2 = \{v_1, v_2,\dots,v_{\omega(G)}\}, T_3 = \{e\}$. Let the root node $r$ to be connected to all nodes in $T_2$. Connect all nodes in $T_2$ other than $v_1, v_2$ to each other. Then connect $v_1$ to all nodes in $T_3 \cup T_2 \setminus \{v_2\}$, and $v_2$ to $T_3 \cup T_2 \setminus \{v_1\}$. Now $d(x) = \omega(G), x \in T_1 \cup T_2$. But now, if $e$ is connected to the required remaining $\omega(G) - 2$ nodes there exists a clique with size $\omega(G) + 1$, which is a contradiction.
Edit: The graph $G$ is an interval graph.
"if $d(v) \geq \omega(G)$, then the subgraph contains a clique larger than the largest clique in the original graph"
Above statement is not true
Counterexample: Please see the below graph. It contains a subgraph $\{1,2,3\}$, where the degree of each node is $4$, i.e. larger than the maximum clique size in the original graph, but the graph G does not have clique larger than size 3.