Rudin 4.22 Theorem Could you help me understand 
why 1. f(H) = B and 
why 2. $\bar A$ $\cap$ B is empty and
why 3. $\bar G$ $\cap$ H is empty?


In 2. (@Antoine's solution) the closure is understood in subspace topology. 
In 3. An indirect proof is (technically) simpler. 
Assume that $x\in \overline G\cap H$. Then $f(x)\in f(\overline G)\subseteq \overline A$, and $f(x)\in f(H)=B$, which is a contradiction because $\overline A\cap B=\emptyset$. In fact, $f(H)\subseteq B$ is enough for the conclusion, that is trivial.
