Let us write $t=(12)$ and $s=(123)$, two elements of the symmetric group $S_3$ of degree $3$.
Construct a directed graph $\Gamma$ as follows:
the vertices are the elements of $S_3$,
if $g\in S_3$ is a vertex, there are two edges coming out of $g$ in $\Gamma$: one going from $g$ to $gt$ and the other going from $g$ to $gs$.
In other words, the set of edges is $E=\{(g,gt)\in S_3\times S_3:g\in S_3\}\cup \{(g,gs)\in S_3\times S_3:g\in S_3\}.$
We can draw a picture:

There is an action of $S_3$ on $\Gamma$ as follows: if $h\in S_3$, then
the action of $h$ on the vertices of $\Gamma$ is by left multiplication by $h$: that is, a vertex $g\in S_3$ is mapped to $hg$;
on the other hand, the action of $h$ on the edges is the induced one: if $(g_1,g_2)$ is one of the edges, then $h\cdot(g_1,g_2)=(hg_1,hg_2)$. It is easy to see that this latter element is, indeed, an edge of $\Gamma$.
It is very easy to see that the action of $S_3$ on the vertices of $\Gamma$ is simply transitive, so that the quotient graph $\Gamma/S_3$ has exactly one vertex, and that the action of $S_3$ on the edges of $\Gamma$ has exactly two orbits. It thus follows that $\Gamma/S_3$ is a two-leaved rose.
$♦ ♦ ♦$
Can you see how to go from this action of $S_3$ on $\Gamma$ to an action of $S_3$ on a CW-complex of dimension $1$, which is what you want?