Is this a valid proof of infinitely many odd integers?
Assume, to the contrary, that there are finitely many odd integers.
Let $S$ be the set of all positive odd integers and let $x=\sum_{n\in S} n$.
Then, $|S|$ is even or $|S|$ is odd.
Let $|S|$ be odd.
Then, $x$ is an odd integer. Let $y=x+2$. Then, $y$ is a positive odd integer not contained in $S$, which is a contradiction.
Let $|S|$ be even.
Then, $x$ is an even integer. Let $y=x+1$. Then, $y$ is a positive odd integer not contained in $S$, which is a contradiction.
Yes, it is a valid proof but I think it is over complicated.
If $S$ is finite then $S$ has a largest member $z$ such that $z \ge x \space \forall x \in S$. Since $z$ is odd then $z' = z+2$ is also odd. But $z \not \ge z' \Rightarrow z' \not \in S$. So we have found a positive odd integer that is not in S. This contradicts the assertion that $S$ is the set of all positive odd integers.
Edit: This is the same as the proof that @JMoravitz outlined in comments above.