How to Show that Points and Closed Sets Can be Separated by Closed Sets in a T3 (Regular) Space 
If $X$ is a regular topological space, given $b \in X$, $A \subset X$
  closed, $b\not\in A$, there are open sets $U$ and $V$ such that
  $\bar{U}\cap\bar{V} = \emptyset$ and $A \subset U, b \in V$.

I was trying to play with the definition. I tried:

Once $X$ is regular then there are $Z$ and $W$ open sets such that $Z
 \cap W = \emptyset,$ $A \subset Z$, and $b \in W.$ 
Then $X - W$ is closed and $b \not\in X - W$. 
Applying again that $X$ is regular, there are $U$ and $V$ such that $X
 - W \subset U$ and $b \in V$, $U \cap V = \emptyset...$

But I can't conclude the result, I don't know if what I have done so far helps somehow... I appreciate any suggestions, answers, hints, etc.
Thanks in advance
 A: Your proof looks correct to me. Just to be clear, I would phrase it is as follows; here I put in bold the subtle points which I justify at the end of the proof.

Because $X$ is regular, we take disjoint open neighborhoods $Y$ and $Z$ containing $A$ and $b$ respectively.
Then define $W=\bar{Y}$ to be the closure of $Y$. Then $W$ is a closed set not containing $b$. $^1$
Now use the regularity property of X to find disjoint open neighborhoods $C$ and $D$ containing $b$ and $W$ respectively.
Since $Y\subset D$, and $W\subset D$, we have $W\cap C=\bar{Y} \cap C = \emptyset$.
Moreover, I claim that $\bar{Y}\cap\bar{C}=\emptyset$. $^2$ 

1. To prove this, we proceed by contradiction. If W contained $b$, then $b \in \partial Y$ (the boundary of $Y$). From this it would follow that every open neighborhood of $b$ would intersect $Y$.
However, $Z$ is an open neighborhood of $b$ which fails this condition. Hence this condition is not true, and we have a contradiction. Therefore $b \not\in W$.
2. If $\bar{Y}\cap\bar{C}\not=\emptyset$, then $W \cap \bar{C} \not= \emptyset$.
However, this contradicts our assumption that $C$ and $D$ are disjoint.
This is because $D-W\not=\emptyset$ and is open, thus $W \cap \bar{C} \not= \emptyset \implies \bar{C} \cap (D-W)\not=\emptyset$. 
If $x \in \bar{C} \cap (D-W)$, then every open neighborhood of $x$ intersects C.
Moreover, since $(D-W)$ is open, then $x$ has a neighborhood $Q \subset (D-W)\subset D$, this neighborhood must also intersect $C$, and this is a contradiction.
(I.e. since $Q \cap C \subset C \cap D$, and $Q \cap C \not= \emptyset$, then $C \cap D \not= \emptyset$.)

Hence finally take $U = Y$ and $V=C$.
