I need to prove that if equation $x = g(x)$ has solution $x^*$ and $g$ is contracting in open interval $(x^* - d, x^*+d)$ where $d > 0$ then for every $x_0$ in $(x^*-d, x^*+d), x_n \mapsto x^*$.
I don't know where to start. I already have a proof for closed interval but it does not apply to open interval. Can someone give me some hint or starting point?
I am not sure how you closed-interval proof works. But would this work? (I am assuming, because your question does not say, that $x_{n+1}=g(x_{n})$.)
Denote $I=(x^{*}-d,x^{*}+d)$. Then there exists unique $x^{*}\in I$ such that $x^{*}=g(x^{*})$ (proof by contradiction assuming two or more solutions exist). By definition of contraction, $k|x-y|>|g(x)-g(y)|$, so, in particular, $k|x_{n}-x^{*}|>|g(x_{n})-g(x^{*})|$, or, equivalently, $k|x_{n}-x^{*}|>|x_{n+1}-x^{*}|$ (you probably need to prove that given any $x_{0}$ the sequence stays in $I$ for the argument to be complete). Hence you have a sequence $\{x_{n}\}$ with the property $k|x_{n}-x^{*}|>|x_{n+1}-x^{*}|$ and hence $k^{n}|x_{0}-x^{*}|>|x_{n+1}-x^{*}|$. Thus $|x_{n+1}-x^{*}|\rightarrow0$.