Lemma
Let $f:X\rightarrow Y$ and $g:Y\rightarrow Z$ function. If $(g\circ f)$ is a bijection then $g$ is surjective.
Proof. So if $(f\circ g)$ is a surjection then by $\text{AC}$ there exist a function $k:Z\rightarrow X$ such that $g\circ (f\circ k)=(g\circ f)\circ k=\text{Id}_Z$ so that $g$ is a surjection.
So if we don't accept $\text{AC}$ does the lemma hold? Indeed if $(g\circ f)$ is a surjection then $Z=(g\circ f)[X]=g\big[f[X]\big]$ but $f[X]\subseteq Y$ so that $Z=g\big[f[X]\big]\subseteq g[Y]\subseteq Z$ so that we conclude $g[Y]=Z$ thus $g$ is surjective. So it is my last argument correct? Could someone help me, please?
Yes, the lemma holds without any form of choice. You also only need that $g \circ f$ is a surjection. Then $g$ will automatically be surjective as well:
Let $z \in Z$. Choose $x \in X$ with $(g\circ f)(x) = z$. Then $g(f(x)) = z$ and thus every element in the codomain of $g$ is attained.