= Solution
\b[As printed, the equivalence needs a transitivity hypothesis.] A group acting trivially on two points has $\mathbb C\Omega=\mathbb C_G\oplus\mathbb C_G$, so $V$ is one-dimensional and irreducible, although the action is not even transitive.
Here is the intended result for a transitive action. The set is then finite, and, for a point stabilizer $H$, its <permutation character> is $\chi=\operatorname{Ind}_H^G1$. <Frobenius reciprocity> gives
$$
\langle\chi,\chi\rangle_G=\langle1,\operatorname{Res}_H^G\chi\rangle_H=\#\{H\hbox{-orbits on }\Omega\}.
$$
The last equality follows because an $H$-fixed vector has constant coefficients on each orbit. There is exactly one trivial constituent, since the action is transitive. Therefore, writing $\chi=1+\psi$ for the character of $V$,
$$
\langle\psi,\psi\rangle_G=\#\{H\hbox{-orbits on }\Omega\}-1.
$$
By <character orthogonality>, $V$ is <irreducible> exactly when this equals $1$. One stabilizer orbit is $\{\omega\}$, so this happens exactly when $H$ is transitive on $\Omega\setminus\{\omega\}$, equivalently when $G$ is <two-transitive>. Moreover $V$ is not trivial, since a transitive action has only one independent fixed vector in the whole permutation module.
The complete correction without assuming transitivity is also precise: \b[$V$ is irreducible exactly for a two-transitive action or the exceptional action on two fixed points]. Indeed for a finite set with $r$ orbits, the fixed subspace of $V$ has dimension $r-1$. If $r>1$ and $V$ is irreducible, $V$ must itself be a one-dimensional trivial module, forcing $|\Omega|=2$ and $r=2$. If $\Omega$ is infinite, $V$ is infinite-dimensional and cannot be irreducible: any nonzero vector generates a submodule of dimension at most $|G|$.
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