Could someone please verify whether I am missing anything important in my proof solution?
For ring homomorphism $\phi:R\to S$, prove that if $R$ is a field and $\phi(R)\neq \left \{0_{S} \right \}$, then $\phi(R)$ is a field.
Let $R$ be a field and let $\phi(R)\neq \left \{0_{S} \right \}$. Then $R$ is a commutative ring with identity and every nonzero element of $R$ is a unit. Since $\phi(R)$ is a subring of $S$, then $\phi(R)$ inherits the commutative property and $1_{S}=\phi(1_{R})\in \phi(R)$, so $\phi(R)$ has identity. Finally, since $uu^{-1}=1_{R}$ for all nonzero $u\in R$, if $0\neq \phi(u)\in \phi(R)$, then $\phi(u)(\phi(u))^{-1}=\phi(u)\phi(u^{-1})=\phi(uu^{-1})=\phi(1_{R})=1_{S}$ Then every nonzero element of $\phi(R)$ is a unit. By definition, $\phi(R)$ is a field.
The last part, I am not sure whether it is logically correct.
Your reasoning looks solid, especially after it takes into account Clayton's (justified) nitpicks.
There are a couple ways of getting this result. My preferred way would be to take advantage of the first isomorphism theorem: $R/\ker(\phi) \cong \text{Im}(\phi)$. Note that $\ker(\phi)$ is an ideal, and the only two ideals of a field are $\{0\}$ and the whole field$^\dagger$. Since $\phi(R) \neq \{0_S\}$, we cannot have $\ker(\phi) = R$. This forces $\ker(\phi) = \{0\}$, from which we can conclude $R \cong \phi(R)$ per the theorem.
$^\dagger$Proof:
Let $F$ be a field and $I \subseteq F$ a nonzero ideal. Then $x \in I$ for some nonzero $x \in F$. Since $F$ is a field, $x^{-1} \in F$, and we'll have $xx^{-1} = 1$ as an element of $I$. Hence, $1 \cdot y \in I$ for all $y \in F \implies I = F$.