The question and my attempts are below.
Let $R \subset S$ be rings, let $\alpha$ be an invertible element of $S$. Show that $R[\alpha] \cap R[\alpha^{−1}]$ is an integral extension of $R$.
Suppose $x \in R[\alpha] \cap R[\alpha^{-1}],$ then $$x=r_{1,n}\frac{1}{\alpha^n}+\dots+r_{1,1}\frac{1}{\alpha^{n-1}} + r_{1,0}\frac{1}{\alpha^0}$$
and $$x=r_{2,n}{\alpha^m}+ \dots +r_{2,1}{\alpha} + r_{2,0}$$
Subtracting these two equations and then multiplication by $\alpha^n$ on both sides gives us a polynomial of order $m+n$,and this shows that $\alpha$ is integral over $R$. This proves the result.
I am still amazed because it if $\alpha$ invertible but not integral over $R$, then the intersection would be empty. Can anyone clear this doubt?
By the last statement, I mean consider $\mathbb{Q} \subset \mathbb{R}$ and consider the irrational number $e \in \mathbb{R}$. Then if any non-zero element $x$ is in the intersection $R[\alpha] \cap R[\alpha^{−1}]$ exists, then it implies $e$ is algebraic. Or there is some mistake in my proof. If there is some mistake, then how should I solve it?
Your proof is not correct. Recall that for $A$ a ring and $K$ a field containing $A$, $x\in K$ is integral over $A$ if there is a monic polynomial $f(X)\in A[X]$ such that $f(x)=0$. Your polynomial is not necessarily monic (it won't be in most cases I think). But more importantly: it is a polynomial with $\alpha$ as root but you need a polynomial with $x$ as root.
The correct proof is different. We use the usual equivalence that $x$ is integral iff there is some finitely generated non-zero $A$-module $M$ stable under $x$-multiplication (i.e. $xM\subseteq M$). You can check that
$$M=A[\alpha^{-n},\dots,1,\dots,\alpha^m]$$
will do the job in your case.
In your proof you confused statements about $x$ and $\alpha$. In fact, you proved a completely different proposition: if $R[\alpha]\cap R[\alpha^{-1}]$ is non-trivially (we always have $R\subseteq R[\alpha]\cap R[\alpha^{-1}]$ but this will show nothing) non-empty, then $\alpha$ is algebraic over $R$. So your example does not work (or does, but you did not realised the implications) as $e\in\mathbb R$ is not algebraic over $\mathbb Q$.