Since $R(3,3)=6$ there is a monochromatic triangle $\Delta$. Let's say it's blue. Look at the other three vertices. If there is no red edge between them then we've found a second blue triangle, so suppose we have found a red edge $xy$, $x,y\notin\Delta$. If there are two blue edges from $x$ to $\Delta$ then we've found a second blue triangle, so assume there are two red edges from $x$ to $\Delta$. Similarly assume there are two red edges from $y$ to $\Delta$. But that means that there is a $z\in\Delta$ such that $xz$ and $yz$ are both red, so we've found a red triangle.
The complete graph $K_{14}$ is much bigger than needed for this result; here are three proper subgraphs with the same property.
I. Every $2$-coloring of the complete bipartite graph $K_{3,7}$ contains a monochromatic $C_4$.
Let $K_{3,7}$ have partite sets $V_1,V_2$ with $|V_1|=3$ and $|V_2|=7$. Each vertex in $V_2$ is joined by edges of one color to two vertices in $V_1$. By the pigeonhole principle, two vertices in $V_2$ are joined by edges of the same color to the same two vertices in $V_1$.
II. Every $2$-coloring of the complete graph $K_6$ contains a monochromatic $C_4$. (This was shown in Misha Lavrov's answer; the following argument is perhaps slightly simpler.)
We may assume that there is a vertex $x$ which is incident with at most two blue edges. In fact, we may assume that $x$ is incident with exactly two blue edges; otherwise $x$ would be incident with four red edges, call them $xa_1$, $xa_2$, $xa_3$, $xa_4$, and then we could consider a vertex $y\notin\{x,a_1,a_2,a_3,a_4\}$ and proceed as in paw88789's answer. (It may be even easier to observe that the subgraph induced by $\{a_1,a_2,a_3,a_4\}$ contains either a red $P_3$ or a blue $C_4$.)
So let $x$ be incident with exactly two blue edges, $xy$ and $xz$. If one of the remaining three vertices is joined by blue to both $y$ and $z$, then we have a blue $C_4$. On the other hand, if each of those three vertices is joined by red to a vertex in $\{y,z\}$, then two of them are joined by red to the same vertex in $\{y,z\}$, and also to $x$, making a red $C_4$.
III. Every $2$-coloring of the complete bipartite graph $K_{5,5}$ contains a monochromatic $C_4$.
The proof is left as an exercise for the reader.
Best Answer
Your proof is OK.
But more easily we can prove more strong and general claim. Assume we have a collection of $n$ irrational numbers. We shall call numbers $a$ and $b$ equivalent if the difference $a-b$ is rational. So we can partition our collection into equivalence classes. We shall call classes $C$ and $C’$ complementary if $c+c’$ is rational for any $c\in C$ and $c’\in C’$. From our partition we can choose such classes which contain in total at least $n/2$ elements and no two complementary classes are chosen. It remains to remark that a sum of any two chosen elements is irrational. In particular, among $5$ irrational numbers we can choose $3$ with all mutual sums are irrational. From the other hand, a collection consisting of $n/2$ numbers $\sqrt{2}$ and $n/2$ numbers $2-\sqrt{2}$ witnesses that the bound $n/2$ is strict.