For any ideal, why must a ring homomorphism be surjective for the ideal to be mapped to an ideal?

Question

Problem:

Let $f$: $R \to S$ be a ring homomorphism

Show that if $f$ is surjective, then for any ideal $I \subset R$, the set $f(I)$ is an ideal of $S$.

My Solution:

Let $x,y\in I\subset R\implies f(x),f(y)\in f(I)\subset S$

1. $0\in I\implies f(0)=0\in f(I)$
2. $x+y\in I\implies f(x+y)=f(x)+f(y)\in f(I)$
3. $-x\in I\implies f(-x)=-f(x)\in f(I)$
4. $r\in R, x\in I, r\cdot x\in I\implies f(r\cdot x)=f(r)\cdot f(x)\in f(I),f(r)\in S, f(x)\in f(I)$
5. $x\in I, r\in R, x\cdot r\in I\implies f(x\cdot r)=f(x)\cdot f(r)\in f(I),f(x)\in I, f(r)\in f(S)$

Thus $f(I)$ is an ideal of $S$

My Question:

If my solution is correct, why does f need to be surjective?