Surjections and Injections
Yes, it is correct, but you should avoid using the word “suggests”. It seems that you are just providing an idea of a proof.
The proof is correct. On the other hand you can avoid contradiction, by proving the contrapositive.
Proposition. Given the maps $f\colon A\to B$ and $g\colon B\to C$, if $g$ is one-to-one and $g\circ f$ is onto, then $f$ is onto.
Proof. Let $b\in B$. Since $g\circ f$ is onto, there exists $a\in A$ such that $$ g(b)=g\circ f(a)=g(f(a)) $$ so $b=f(a)$ because $g$ is one-to-one. Therefore $f$ is onto.$\quad\square$
Your proof is well written and correct.
I would suggest you avoid writing $(z,\alpha)\in f$, however, because you already use the notation of $f(z)=\alpha$, and mixing the two notations makes the whole thing much harder to read.
In fact, the entire last paragraph can be written more cleanly:
Assume that $g\circ f$ is onto. Then there exists some $z$ such that $(g\circ f)(z)=c$. However, because $(g\circ f)(z)=g(f(z))$, we now know $g(f(z))=c$. Define $\alpha=f(z)$, therefore $g(\alpha)=c=g(x)$.
From $(2)$, we can conclude that $\alpha = x$, and because $\alpha=f(z)$, we conclude that $f(z)=x$. This contradicts with $(1)$. Therefore, the original assumption was incorrect.
Conclusion: $g\circ f$ is not onto.