Proving $\sqrt{2}$ is irrational

Part of the problem is that you have written the proof in a stream-of-consciousness style; proofs of subclaims are mixed in with the main proof, and variables are used without any explanation of what they stand for. Let me rewrite the proof so that it is easier to understand.

Proof: Let $S = \{x\in\mathbb{R} \mid \text{for some }a,b\in\mathbb{Z}, x = a+\sqrt{2}b\}$.

Claim: $S$ is closed under multiplication.

Proof of claim: Suppose $x_1,x_2\in S$. Then there are integers $a_1$, $b_1$, $a_2$, $b_2$ such that $x_1 = a_1+\sqrt{2}b_1$ and $x_2 = a_2+\sqrt{2}b_2$. Therefore $$ x_1x_2 = (a_1+\sqrt{2}b_1)(a_2+\sqrt{2}b_2)= (a_1a_2+2b_1b_2) + \sqrt{2}(b_1a_2+b_2a_1)\in S. $$ This proves the claim.

Hence if $x \in S$ then $x^n \in S$ for every $n \in \mathbb{N}$.

Suppose $\sqrt{2}$ is rational. Then there are positive integers $p$ and $q$ such that $\sqrt{2} = p/q$.

Claim: For all $x \in S$, $qx \in \mathbb{Z}$.

Proof of claim: Suppose $x \in S$. Then there are integers $a$ and $b$ such that $x = a+\sqrt{2}b = a+(p/q)b$. Therefore $qx = qa+pb \in \mathbb{Z}$. This proves the claim.

Now let $x = \sqrt{2}-1 \in S$. Note that $0 < x < 1$, so $1/x > 1$. Therefore $\lim_{n \to \infty} (1/x)^n = \infty$. Choose $n \in \mathbb{N}$ with $(1/x)^n > q$, so $0<qx^n<1$. But by the claims above, $x^n \in S$ and therefore $qx^n \in \mathbb{Z}$. This is a contradiction, because there are no integers between 0 and 1.