I am having trouble finding the Ramsey number of $2\times$(complete graph of $2$ vertices) and $3\times$(the complete graph of two vertices).
I don't really understand if $2K_2$ is connected or not, and same with $3K_2$.
$2K_2$ is two disjoint copies of the complete graph $K_2$. $2K_3$ is two disjoint copies of the complete graph $K_3$.
We prove that $R(2K_2,3K_2)=7$.
In a red-blue (edge-)colored $K_6$, let the red graph be $K_{1,5}$. Since all edges are incident with the same vertex, the red graph does not contain $2K_2$. The blue graph is $K_5+K_1$, which does not contain $3K_2$. This proves that $R(2K_2,3K_2)>6$.
Now consider a red-blue colored $K_7$ that does not contain a red $2K_2$ or a blue $3K_2$.
If the red graph contains a cycle, it must be $C_3$, since every longer cycle contains $2K_2$. But then the red graph cannot contain any other edge, since such edge $e$ is incident with at most one of the vertices of the $C_3$ and then we find $2K_2$: the edge between the other two vertices of the $C_3$ is independent of $e$. The blue graph ($K_7-C_3$) clearly contains $3K_2$, contradiction.
So the red graph must be a tree, and because it cannot contain two independent edges, this tree must be a star. The "worst" star on 7 vertices is $K_{1,6}$, so the blue graph certainly contains $K_7-K_{1,6}$ as a subgraph, which contains $C_7$, which contains $3K_2$. Contradiction.
This shows that $R(2K_2,3K_2)\leq7$ and we are done.