How to prove this conjecture:
For any prime $p_{n}>7$, there is at least as much odd composites before $p_{n}$ than between $p_{n}$ and $p_{n+1}$.
Let $o_{n}$ denote the number of odd integers less than or equal to $p_{n}$, exluding $1$:
$$o_{n}=\left\lceil \frac{p_{n}}{2}\right\rceil -1$$
Let $c_{n}$ denote the number of odd composites less than or equal to $p_{n}$:
$$c_{n}=o_{n}-\left(n-1\right)=\left\lceil \frac{p_{n}}{2}\right\rceil -n$$
If $x_{n}$ is the maximum number of odd composites between $P_{n}$ and $P_{n+1}$, then:
$$p_{n+1}-p_{n}=g_{n}\leq2x_{n}+2$$
If $x_{n}\leq c_{n}$, then $g_{n}\leq2c_{n}+2$, so to prove the conjecture we need to prove that:
$$g_{n}\leq2\left(\left\lceil \frac{p_{n}}{2}\right\rceil -n+1\right)$$
The statement of which the OP is seeking a proof is stronger than Bertrand's postulate.
If you rearrange the inequality, you get $2p_n+3\geq p_{n+1}+2n$.
So there isn't really a short, snappy proof of this. You'll have to use some machinery.
The more machinery you use, the more trivial the question.