I think that this is a proof that all even numbers are sum of two odd primes, but would be grateful for anyone who could have a look to see if I missed something
A formal proof will go as follows:- We take two odd primes p1 and p2 So that $p1≥ p2$ $p1+ p2≥ 2p2$ But $p1$ and $p2$ are odd primes and sum of $2$ odd no is always even So $p1+p2=2x$ $2x≥2p2$ $x≥p2 ...(1)$ Now if $p1≥p2$ $p1≥x ...(2)$ combining $(1)$ and $(2)$ $p1≥x≥p2$ Now $p2$ is the smaller prime and $3$ is the smallest prime and there are infinitely many primes so greatest prime is infinite $∞ ≥ x ≥ 3$ So $x$ belongs to $[3,∞]$ , which in turn makes $2x$ every even no starting from $6$. But if sum of two primes is equal to $2x$ then $x$ can also be equal to $2$ $x$ belongs to $2$ to infinity , So $2x$ is every even no greater than $2$. But $2x$ is the sum of two primes $p1$ and $p2$. Every even no greater than $2$ is the sum of two primes. Second case can easily be proved by taking $3,5..$ as the third prime.