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.
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
Lemma: Let $K_{m, n}$ be the complete bipartite graph on sets of vertices $A$ and $B$ with respective sizes $m$ and $n$. If the edges of $K_{m, n}$ are two-colored, then there is a monochromatic connected subgraph spanning at least $\frac{m+n}{2}$ vertices.
Proof: Call the colors red and blue. Call a vertex red-primary if it has more red than blue edges, blue-primary if it has more blue than red edges, and neutral if the numbers are equal. There are two cases.
Case 1. For some set $X \in \{A, B\}$ and for some color $C \in \{\text{red}, \text{blue}\}$, at least one vertex in $X$ is $C$-primary, and at least $|X|/2$ vertices in $X$ are either $C$-primary or neutral.
Suppose without loss of generality that $A$ has at least one red-primary vertex, and the majority of vertices in $A$ are red-primary or neutral. If $a_1 \in A$ is red-primary and $a_2$ is either red-primary or neutral, then by the pigeonhole principle, some vertex in $B$ has red edges to both $a_1$ and $a_2$. Thus, there is a red connected graph spanning at least any red-primary or neutral vertex in $A$ (which makes up at least half of $A$) and any vertex in $B$ with red edges to any of these vertices (which makes up at least half of $B$, as any single red-primary or neutral vertex has red connections to at least half of the other set).
Case 2. Every vertex is neutral. Then any vertex $A$ has red connections to exactly half the vertices of $B$. Each of those vertices has a red connection to exactly half of $A$ (perhaps a different half each time). This gives us a red monochromatic subgraph.
Now to the main problem. The cases $n = 1, 2, 3, 4$ can be established trivially. The required subgraph for $n = 2m-1$ odd is has the same size as that for $n = 2m$ even, so it suffices to prove the case for odd $n$.
We work by induction, showing that the claim for $n=2m-1$ implies the claim for $n=2m+1$. Let the colors be red, blue, and green, and let $K_{2m+1}$ be a complete edge-3-colored graph on $2m+1$ vertices. By the induction hypothesis, $K_{2m+1}$ has a connected monochromatic (without loss of generality, green) subgraph of $m$ vertices. Let $G$ be this set of vertices, and let $H$ be the rest of $K_{2m+1}$. If any of the edges from $G$ to $H$ is green, then we have a connected green graph of the required size $m+1$. Otherwise, erasing every edge internal to $G$ or to $H$ gives a complete bipartite graph on $m$ and $m+1$ vertices with every edge colored either red or blue, and the lemma guarantees either a red or a blue monochromatic subgraph with the required size.
Remark: It is possible to construct instances in which the given bound is tight. For example, for $n = 4m$ even, one may construct a graph with no monochromatic graph larger than $2n$ by taking four equally sized sets of vertexes $A, B, C, D$ and coloring all edges internal to the sets $A \cup B$ and $C \cup D$ green, edges from $A$ to $C$ or $B$ to $D$ red, and edges from $A$ to $D$ or $B$ to $C$ blue.