Given $\epsilon>0$ and $0<\delta\leq\min\{1,\sqrt{\epsilon}\}\,$, we have that $\,|x|<\delta\implies0<x^2<1\implies0<\left|\frac{1}{x^2-1}\right|<1$, thus $$\left|\frac{1}{x^2-1}+1 \right|=\left|\frac{x^2}{x^2-1}\right|=\left|\frac{1}{x^2-1} \right|\cdot|x|^2<1\cdot\left(\sqrt{\epsilon}\right)^2=\epsilon.$$
Is this proof correct?
No. If $0<x^2<1$ then $-1<x^2-1<0$ so $$\frac1{x^2-1}<-1$$ so $$\Bigl|\frac1{x^2-1}\Bigr|>1\ .$$ You will need to revise your inequalities (and probably also your choice of $\delta\,$).
Other small points: you should say $$\delta=\langle\hbox{something}\rangle\quad\hbox{and}\quad 0<|x|<\delta\ ,$$ not $\delta\le\cdots$ and $|x|<\delta$.