Prove $S/\operatorname{Ker}(\alpha) \cong \operatorname{Im}(\alpha)$ where $\alpha: S \rightarrow T$ is a ring homomorphism.
Proof: Let $A = S/\operatorname{Ker}(\alpha)$, $B = \operatorname{Im}(\alpha)$
Then $\alpha$ induces a map $\tilde{\alpha}: S/\operatorname{Ker}(\alpha) \rightarrow \operatorname{Im}(\alpha)$. This map is given by: $\tilde{\alpha}(\tilde{x})=\alpha(x)$.
To show that this map is well defined, suppose $\tilde{y} \in \operatorname{Ker}(\alpha)$, then we have:
$\tilde{\alpha}(\tilde{x}+\tilde{y})=\alpha(x+y)=\alpha(x)+\alpha(y)=\alpha(x)+0=\alpha(x)$
Let $\tilde{x}, \tilde{y} \in S/\operatorname{Ker}(\alpha)$ and suppose $\tilde{\alpha}(\tilde{x})=\tilde{\alpha}(\tilde{y})$. This implies that $\alpha(x)=\alpha(y)$, and thus $\alpha(x-y)=\alpha(x)-\alpha(y)=0$.
Thus $x-y \in \operatorname{Ker}(\alpha)$ and so $\tilde{x}=\tilde{y}$. Thus $\tilde{\alpha}$ is one-to-one.
Now let $z \in \operatorname{Im}(\alpha)$, then $\exists s \in S$ such that $\alpha(s)=z$. This implies that $s \neq 0$ and so $\tilde{\alpha}(\tilde{s})=z$.
Thus $\tilde{\alpha}$ is bijective, and it is also a ring homomorphism as it was induced by a ring homomorphism. Therefore $\tilde{\alpha}$ is an isomorphism.
Q.E.D.
How does this look? I feel like there is a lot I could make more explicit / sharpen up.
As I cannot comment yet, I'll give you my thoughts here:
One thing you did not mention is the definition of the induced morphism $\bar{\alpha}$. Then you would also need to check, that this definition, which depends on choosing a representative, is well defined.
In your proof of surjectivity, you don't get that $s\neq 0$. But this is not needed anyway.
Apart from that your proof is complete and correct.