Proving that the integral closure of $R$ in $S$ is a subring of $S.$

You prove that it is a subring by ordinary "subring test" but you use the lemma while doing so. Although in our commutative algebra course we used also the following version of lemma (we actually prove this first to get your lemma but I am not sure whether it is necessary to do it in this order. )

Lemma 1 $Rsubset S$, $xin S $ then $x$ is integral over $R$ if $R[x]$ is finitely generated $R-$module.

Lemma 2 $x$ is integral over $R$ iff there exists finitely generated $R-$module $R'$ such that $Rsubset R[x]subset R'subset S$

Take $a,bin S$ integral over $R$. By your lemma $R[a]$ is finitely generated $R$-module. But we can view $R[a]$ simply as a ring so $b$ is integral also over $R[a]$. Hence $R[a][b]$ is finitely generated module over $R[a]$.

Now $R[a][b]=R[a,b]$ as subrings of $S$ (they consist of the same elements) so $R[a,b]$ is also finitely generated as $R-$module.

Using lemma 2: Now set $x:=a+b$ and $R':=R[a,b]$. $R'$ is finitely generated $R-$module and $Rsubset R[x]subset R'subset S$ so by lemma 2 we conclude that $x$ is integral.

Commentary I think that the reason we need to use lemma 2 instead of ypur version (lemma 1) is that you can't conclude that $R[a+b]$ is finitely generated from the fact that it is $R-$submodule of finitely generated $R-$module $R[a,b]$. there are submodules of fin. gen. modules that are not finitely generated (take ring $R$ as a module over itself it is generated by $1$ but unless it is Noetherian it has ideals that are not finitely generated)

However, in this particular setting, lemma 2 gives you that $a+b$ is integral so by lemma 1 is $R[a+b]$ finitely generated $R-$module after all.