Could someone please verify whether my solution is correct?
For ring homomorphism $\phi:R\to S$, prove that $\phi(R)$ is a subring of $S$.
Subset/nonempty: Since $\phi(0_{R})=0_{S}$, then $\phi(R)\subset S$ and $\phi(R)\neq \varnothing$.
Closure under subtraction: Let $\phi(r),\phi(s)\in \phi(R)$. Since $R$ is a ring, $r-s\in R$ and $\phi(r-s)\in S$. Then $\phi(r-s)=\phi(r)-\phi(s)\in \phi(R)$.
Closure under multiplication: Since $rs\in R$ and $\phi(rs)\in S$, then $\phi(rs)=\phi(r)\phi(s)\in \phi(R)$.
I am not sure what to say for closure, whether it should be stated that $\phi(rs)\in S$, etc.
$\phi(R)\subseteq S$. To show it's a subring, we use the subring test: closure under subtraction and multiplication.
Pick $s_1,s_2\in\phi(R)$. There will be $a_1,a_2\in R:s_1=\phi(a_1),s_2=\phi(a_2)$. But then $s_1-s_2=\phi(a_1)-\phi(a_2)=\phi(a_1-a_2)\in\phi(R)$, so closure under subtraction is done.
Pick $s_1,s_2\in\phi(R)$. Again, we have the above $a_1,a_2$. But then $s_1s_2=\phi(a_1)\phi(a_2)=\phi(a_1a_2)\in\phi(R)$, and we're done.