[proofplan]
Fix an arbitrary vertex $x \in V(K_6)$. Since $x$ is incident to five edges coloured red or blue, by the pigeonhole principle at least three of these edges share a colour — say red. Let $y_1, y_2, y_3$ be the other endpoints of three such red edges. Now inspect the three edges among $\{y_1, y_2, y_3\}$: if any one is red, it closes a red triangle with $x$; otherwise all three are blue and form a blue triangle among $\{y_1, y_2, y_3\}$. Either way a monochromatic triangle exists.
[/proofplan]
[step:Fix a vertex and extract a monochromatic star of size three by pigeonhole]
Fix any $x \in V(K_6)$. Since $K_6$ is complete on six vertices, $x$ is adjacent to the five remaining vertices via five edges. The colouring $c : E(K_6) \to \{\text{red}, \text{blue}\}$ assigns each of these five edges one of two colours. By the pigeonhole principle, at least $\lceil 5/2 \rceil = 3$ of them share a common colour. Relabelling colours if necessary (the problem is symmetric in red and blue), we may assume without loss of generality that three of the edges at $x$ are red, with other endpoints $y_1, y_2, y_3 \in V(K_6) \setminus \{x\}$. Thus
\begin{align*}
c(x y_1) = c(x y_2) = c(x y_3) = \text{red}.
\end{align*}
[guided]
The strategy is to reduce the six-vertex problem to a three-vertex sub-problem. How? Pick any vertex $x$ and use the fact that it has five neighbours, coloured with only two colours, so the majority colour appears at least three times.
Explicitly: $x$ is adjacent to each of the remaining $|V(K_6) \setminus \{x\}| = 5$ vertices via a single edge, and the colour of each edge lies in $\{\text{red}, \text{blue}\}$. If fewer than three edges were red and fewer than three were blue, we would have at most $2 + 2 = 4$ edges — but there are $5$, contradiction. So one colour appears at least three times at $x$.
Why WLOG red? The statement of the theorem is symmetric in the two colours: swapping red and blue in any colouring sends monochromatic red triangles to monochromatic blue triangles and vice versa. So we may relabel and assume the majority colour at $x$ is red. Let $y_1, y_2, y_3$ be three vertices joined to $x$ by red edges:
\begin{align*}
c(x y_1) = c(x y_2) = c(x y_3) = \text{red}.
\end{align*}
[/guided]
[/step]
[step:Case-split on the colours of the three edges among $\{y_1, y_2, y_3\}$]
Consider the three edges $y_1 y_2$, $y_1 y_3$, $y_2 y_3$ inside the triangle on $\{y_1, y_2, y_3\}$.
**Case 1.** At least one of these three edges is red. Say $c(y_i y_j) = \text{red}$ for some $i \neq j$ in $\{1, 2, 3\}$. Then the three vertices $x, y_i, y_j$ span the edges $x y_i$, $x y_j$, $y_i y_j$, all of which are red. Hence $\{x, y_i, y_j\}$ is a red triangle.
**Case 2.** None of the three edges is red. Since the only other colour is blue, all three edges $y_1 y_2$, $y_1 y_3$, $y_2 y_3$ are blue. Hence $\{y_1, y_2, y_3\}$ is a blue triangle.
In both cases a monochromatic triangle exists.
[guided]
We have identified three vertices $y_1, y_2, y_3$ each joined to $x$ by a red edge. The remaining unknown structure is the colouring of the triangle on $\{y_1, y_2, y_3\}$, which consists of three edges: $y_1 y_2$, $y_1 y_3$, $y_2 y_3$. We classify by whether any of these is red.
**Case 1 — some edge $y_i y_j$ is red.** The three vertices $\{x, y_i, y_j\}$ form a triangle in $K_6$ with edges $x y_i$, $x y_j$, $y_i y_j$. We know $c(x y_i) = c(x y_j) = \text{red}$ by our choice of $y_1, y_2, y_3$, and $c(y_i y_j) = \text{red}$ by the case hypothesis. All three edges are red, so this triangle is monochromatically red.
**Case 2 — no edge $y_i y_j$ is red.** Every edge among $\{y_1, y_2, y_3\}$ is coloured red or blue; since none is red, all three are blue:
\begin{align*}
c(y_1 y_2) = c(y_1 y_3) = c(y_2 y_3) = \text{blue}.
\end{align*}
Thus the triangle $\{y_1, y_2, y_3\}$ is monochromatically blue.
Either case produces a monochromatic triangle, proving the theorem.
[/guided]
[/step]
