I found the claim in a an article , here is my proof attempt:
Let $G$ be a primitive group (transitive) on a set $\Omega$ of $n$ elements.
By the definition of doubly transitive, there exists an element $g \in G$ such that $(a, b)^g = (c, d)$ where distinct $a, b,c,d \in \Omega$.
Since there are $n(n-1)$ distinct $(a, b)$ tuples, if we assign an element/permutation of $G$ to each distinct $(a, b)$ tuple satisfying the definition of doubly transitive, there should be $n(n-1)$ elements in $G$, but $n(n-1)$ is even, thus we prove the claim.
Is this proof correct? Is there an alternative proof?
Proceed by contradiction and assume $(G,\cdot)$ is a doubly transitive permutation group on $\Omega$ but $2\nmid|G|$ where $|\Omega|>1$. By the double transitivity of $G$ we may fix distinct $a,b\in\Omega$ and $\pi\in G$ such that $\pi(a)=b$ and $\pi(b)=a$. But $(a\;b)$ is a $\pi$ cauchy cycle and thus Ruffini's theorem guarantees $2\mid\text{ord}(\pi)$ which in turn implies $2\mid|G|$ via Lagrange's theorem. $\it{reductio\;ad\;absurdum}$.