Here's a proof, from Wikipedia, that $K_6$ has at least 2 monochromatic triangles. You may find that this method of proof generalizes to give you the lower bound you want.
"An alternate proof works by double counting. It goes as follows: Count the number of ordered triples of vertices $x, y, z$ such that the edge $(xy)$ is red and the edge $(yz)$ is blue. Firstly, any given vertex will be the middle of either $0\times5=0$ (all edges from the vertex are the same color), $1\times4=4$ (four are the same color, one is the other color), or $2\times3=6$ (three are the same color, two are the other color) such triples. Therefore there are at most $6\times6=36$ such triples. Secondly, for any non-monochromatic triangle $(xyz)$, there exist precisely two such triples. Therefore there are at most 18 non-monochromatic triangles. Therefore at least 2 of the 20 triangles in the $K_6$ are monochromatic."
On the other hand, if you want all the work done for you, there is an answer in A W Goodman, On sets of acquaintances and strangers at any party, Amer. Math. Monthly 66 (1959) 778-783, also in Frank Harary, The two-triangle case of the acquaintance graph, Math. Mag. 45 (1972) 130-135.
You can simplify(?) your argument by using induction:
Let $f(k)=R(\underbrace{3,3,\ldots,3}_k)$.
In any $k$-colouring ($k\ge1$) of a $K_n$ graph ($n\ge2$), if we pick any vertex $v$, then by the pigeon-hole-principle, one of the $k$ colours ("blue", say) will occur for at least $m:=\left\lceil\frac{n-1}{k}\right\rceil$ times among the edges incident with $v$. If any edge between the other ends of these edges is blue, we have a blue triangle and are done. And if none of these edges is blue, we have found a subgraph that is a $(k-1)$-coloured $K_m$. Hence if $m\ge f(k-1)$ we are done again.
In other words, we almost immediately find a monochromatic triangle if $\frac {n-1}k> f(k-1)-1$, or equivalently if $n>kf(k-1)-k+1$. We conclude that $$f(k)\le kf(k-1)-k+2.$$
Clearly, $f(1)=3$. From this, $f(2)\le 6$, $f(3)\le 17$, $f(4)\le 66$, etc.
Thus we have found an upper bound
$$ f(k)\le a_k$$
where $a_k$ is defined by the recursion $$\begin{align}a_1&=3,\\ a_{k}&=ka_{k-1}-k+2\qquad\text{for }k\ge2.\end{align}$$
Claim. We have $a_k-1=\sum_{i=0}^k\frac{k!}{i!}$.
Proof.
This is clear for $k=1$.
For $k\ge 2$ we find by induction
$$a_k-1=k(a_{k-1}-1)+1=k\sum_{i=0}^{k-1}\frac{(k-1)!}{i!}+\frac{k!}{k!}=\sum_{i=0}^k\frac{k!}{i!}$$
$\square$
Corollary.
$$R(\underbrace{3,3,\ldots,3}_k)\le \left\lceil k!e\right\rceil$$
Proof. This follows from $e=\sum_{i=0}^\infty\frac1{i!}$ and that the error $\sum_{i=k+1}^\infty\frac1{i!}$ is between $0$ and $1$ for $k\ge1$. $\square$
Remark. See also A073591.
Best Answer
Lemma: Any two coloring of $K_6$ produces a monocromatic triangle.
Note: This does not hold necessery for $K_5$. Find an example.
So if we have a colored $G\simeq K_6$ then we are done. So if in $K_9$ there is no $G$ then we have by Turan's theorem:
$$\varepsilon \leq \Big[{2\cdot 9^2\over 5}\Big] =32$$
So if $K_9$ has at least $33$ colored edges there is colored $K_6$ and thus by lemma a conclusion.
Construction of nonexistence for $32$ edges is easy: Make 4 sets $A_1,A_2,A_3,A_4$ each with two vertices and $A_5$ with a single vertex.
Then connect all vertices between $A_i$ and $A_j$ with blue edges iff $|i-j|=1$ modulo $5$ and else with red edges. Edges in each $A_i$ are uncolored.