TITLE: Any partition of $\{1,2,\ldots,9\}$ must contain a $3$-Term Arithmetic Progression
QUESTION [5 upvotes]: Prove that for any way of dividing the set $X=\{1,2,3,\dots,9\}$ into $2$ sets, there always exist at least one arithmetic progression of length $3$ in one of the two sets.

REPLY [6 votes]: (This is very similar to Andres Caicedo's proof, but is a little simpler.)
Let us call the sets red and blue.  Consider the elements $\{4,5,6\}$.  At least two of these are the same color;  say red.  Then we have three cases: $\{4,5\}$ are red, $\{4,6\}$ are red, or $\{5,6\}$ are red.  $\def\r#1{\color{red}{#1}}\def\b#1{\color{blue}{#1}}$

$\r4$ and $\r5$ are both red.  Then if $3$ or $6$ is red we are done, so suppose $\b3$ and $\b6$ are blue.  Then $\r9$ is red (else $\b3,\b6,\r9$ is blue), so $\b1$ is blue (else $\b1,\r5,\r9$ is red), $\r2$ is red ($\b1,\r2,\b3$), $\b8$ is blue ($\r2,\r5,\b8$).  Now if $7$ is red we have $\r5,\r7,\r9$ and if $7$ is blue we have $\b6,\b7,\b8$.
$\r4$ and $\r6$ are both red.  Then $\b2$ and $\b5$ are both blue (else $\b2,\r4,\r6$ or $\r4,\b5,\r6$).  Now if $8$ is red we have red $\r4,\r6,\r8$ and if $8$ is blue we have blue $\b2,\b5,\b8$. 
$\r5$ and $\r6$ are both red.  This reduces to  the $\{4,5\}$ case by the transformation $x\mapsto (10-x)$.