$R_0$ and normal implies completely regular We say a space ($X, \tau$) is $R_0$ if for each $x,y \in X$ such that $\overline{\{x\}}\neq\overline{\{y\}}$ there exists open sets with $x\in U-V$ and $y \in V-U$.
If for each $F, K \subseteq X$ disjoint closed sets there exists disjoints open sets $U$ and $V$ such that $F \subseteq U, K \subset V$ we say that the space is normal.
And if for each $x \in X$ and $F \subset X$ closed such that $x \notin F$ there exists $f:X \rightarrow [0,1]$ continuous with $f(x)=0$ and $f[F]=\{1\}$ we say that the space is completely regular.
I want to show:

Theorem: If a space is $R_0$ and normal, then is completely regular.

I've tried to show that with this conditions the space must be $T_1$.  If  $\overline{\{x\}} \neq \overline{\{y\}}$ for any pair of distinct points then I'm done, this is equivalent to be $T_0$, but I don't know if this is always true?
 A: The space need not be $T_1$. Let $X=\{0,1,2\}$ and $\tau=\big\{\varnothing,\{0\},\{1,2\},X\big\}$; then $\langle X,\tau\rangle$ is $R_0$ and normal but not $T_0$ (and hence not $T_1$). And it is completely regular but not Tikhonov.
HINT: To prove the theorem, use Urysohn’s lemma: if $F$ and $K$ are disjoint closed set in $X$, there is a continuous $f:X\to[0,1]$ such that $f[F]=\{0\}$ and $f[K]=\{1\}$. Suppose that $x\in X$, $F\subseteq X$ is closed, and $x\notin F$. Let $K=\operatorname{cl}\{x\}$. If $K\cap F=\varnothing$, we can immediately apply Urysohn’s lemma to get the desired result.
If not, let $y\in K\cap F$. $X\setminus F$ is an open nbhd of $x$ that does not contain $y$, so $x\notin\operatorname{cl}\{y\}$, and therefore $\operatorname{cl}\{x\}\ne\operatorname{cl}\{y\}$. Now use the hypothesis that $X$ is $R_0$ to get a contradiction, showing that in fact $K\cap F=\varnothing$.
A: (Brian has already provided an answer in the form of hints, which is what the OP initially requested. Here I'll rewrite it community wiki as a full solution so users can have it for reference.)
Lemma: Let $X$ be an $R_0$ space.  If $F$ is a closed set and $x\in X$ is a point not in $F$, then $F\cap\overline{\{x\}}=\emptyset$.
Proof: Suppose there is some $y\in F\cap\overline{\{x\}}$.  Since $F$ is closed, $\overline{\{y\}}\subseteq F$, and hence $x\notin\overline{\{y\}}$.  So $\overline{\{x\}}\ne\overline{\{y\}}$.  By definition of $R_0$ there is a nhbd of $y$ not containing $x$.  But that implies $y\notin\overline{\{x\}}$, contradiction.
Theorem: Suppose the space $X$ is $R_0$ and normal.  Then $X$ is completely regular.
Proof: Given a closed set $F$ and a point $x\notin F$, the sets $F$ and $\overline{\{x\}}$ and closed and disjoint by the lemma.  Urysohn’s lemma applied to these two sets then provides a continuous function $f:X\to[0,1]$ with $f[F]=\{1\}$ and $f(x)=0$.  This shows that $X$ is completely normal.
