Here is the question above. I am stuck in part a) .
here I am able to find a positive $f\in J$ with $\|f\|_\infty \le 1$ such that $f\ne 0$ on $X-U$ to get $f=1$ on $X-U$ we need to extend $1/f$ on $X$ which can be done by Teitze-extension . But how to control the norm of $g$ such that $\|f.g\|_\infty\le1$ where $g$ is the extension of $1/f$.
Let a function $f\in J$ be positive on $X\setminus U.$ Thus $f$ is bounded below on $X\setminus U$ by a positive constant $c$. Replacing $f$ with $c^{-1}f$ we may assume that $f$ is bounded below by $1.$ Let $$g(x)=\begin{cases}f(x)& f(x)\le 1 \\ \displaystyle {1\over f(x)} & f(x)>1\end{cases}$$ Then $g$ is continuous. Moreover $fg\in J$ and $$g(x)f(x)=\begin{cases}f(x)^2& f(x)\le 1 \\ \displaystyle {1} & f(x)>1\end{cases}$$ Hence $0\le fg\le 1$ and $f(x)g(x)=1$ for $x\in X\setminus U. $