Show that there exists a set $U$ which is both open and closed and $x \in U \subseteq V$. 
Let $X$ be a compact topological space. Suppose that for any $x, y \in X$ with
  $x \neq y$, there exist open sets $U_x$ and $U_y$ containing $x$ and $y$, respectively, such that $$ U_x \cup U_y = X\quad \text{and}\quad U_x \cap U_y = \varnothing.$$
  Let $V \subseteq X$ be an open set. Let $x \in V$ . Show that there exists a set $U$ which is both open and closed and $x \in U \subseteq V$.

My Try:
$\forall \;x\neq y\quad U_x \cap U_y = \varnothing$, $X$ is Hausdorff hence a $T_1$-space. $X$ is compact Hausdorff hence normal subsequently normal $T_1$-space i.e. $T_4$-space. As a consequence of $T_1$-space, $\{x\}$ is closed in $X$.
Given $V$ is open and $x\in V$. So $\bbox[5px,border:1px solid red]{\text{there exists open set $U$ such that  $x \in U \subseteq V.$}}$ $U^c$ is closed. By normality of $X$ there exists disjoint open sets $W_1$ and $W_2$ such that 
$$\{x\}\subseteq W_1\subseteq U \;\text{and}\; U^c\subseteq W_2\implies W_2^c\subseteq U\tag 1$$
$$W_2^c\subseteq U\implies X=W_2\cup W_2^c\subseteq U \cup W_2 \tag 2$$
Claim: $U\cap W_2=\varnothing$.
For suppose $x\in U\cap W_2$ then,
$$x\in U \;\text{and}\; x\in W_2\implies\bbox[5px,border:1px solid red]{{ x\not\in U^c \;\text{or}\; x\not\in W^c_2\implies x\not\in U^c\cup W^c_2}}$$
Also from $(1)$ and $(2)$ we see that $$U^c\cup W^c_2\subseteq U\cup W_2\implies x\not\in U^c\cup W^c_2\subseteq U\cup W_2$$
a contradiction since $x\in U\cap W_2$ but $x\not\in U \cup W_2$. Hence the claim subsequently $U$ is both open and closed such that $x \in U \subseteq V$.
Is there anything incorrect or missing in my proof ? Are there any other alternative proofs?

Update: I figured the two highlighted portions are the incorrect parts of the proof. Thanks to @Thomas Andrews and @Hagen von Eitzen for clarifying my doubts and for their answers.
 A: We are given (with improved notation) that for $x,y\in X$ with $x\ne y$, there exist open sets $U_{(x,y)}$ and $U_{(y,x)}$ such that
$$x\in U_{(x,y)},\quad y\in U_{(y,x)},\quad U_{(x,y)}\cup U_{(y,x)} =X,\quad U_{(x,y)}\cap U_{(y,x)} =\emptyset. $$
Now assume $x\in X$ and $V\ni x$ is open.
Then $V$ together with all $U_{(y,x)}$, $y\in V^\complement$ form an open cover of $X$. Pick a finite sub-cover consisting of $V$ and some $U_{(y_i,x)}$, $i=1,2,\ldots, n$.
Let $$U=\bigcap_{i=1}^nU_{(x,y_i)}.$$
Then $U$ is a finite intersection if clopen sets, hence clopen. Clearly $x\in U$. And as the $U_{(y_i,x)}$ cover $V^\complement$, it follows that $U\subseteq V$.
A: The problem in your proof is after you conclude: $$x\not\in U^c\cup W^c_2\subseteq U\cup W_2$$
From which you deduce, incorrectly, that $x\notin U\cup W_2.$ But if $x\notin A$ and $A\subseteq B$ you cannot conclude $x\notin B.$ Just take, for example, $B=A\cup \{x\}.$

A proof. 
For each $y\notin V,$ we have a both closed and open set $U_y$ with $x\notin U_y.$ Then the set of all $U_y$ is an open cover of $V^c,$ a closed subset of compact space, so there must be $y_1,\cdots, y_n$ which cover $V.$ 
But then $U=U_{y_1}^c\cap U_{y_2}^c\cap \cdots \cap U_{y_n}^c$ is a finite intersection of sets that are both closed and open, and hence $U$ is both closed and open. Also, since $x\in U_y^c$ for every $y,$ $x\in U$, and finally, since the $U_{y_i}$ cover $V^c,$ we have that $$U=\left(U_{y_1}\cup U_{y_2}\cup \cdots U_{y_n}\right)^c\subseteq V.$$

As noted in a comment to another proof, you have to take some liberties with the argument when $V=X,$ when necessarily $n=0.$ 
If we think of $U$ as: $$U=X\setminus\left(U_{y_1}\cup \cdots \cup U_{y_n}\right)$$ it is more obvious that when $V=X$, and hence $n=0,$ we get $U=X.$
