Prove that $x\not\in P$ then there exists a neighborhood $O$ of $x$ such that $O\cap P=\varnothing$. Certain definition on $P$. I'm reading Intro to Topology by Mendelson.
The problem statement is,

Let $\{U_\alpha\}_{\alpha\in I}$ be an open covering of $[0,1]$. Define a subset $P$ of $[0,1]$ as follows:
$x\in P$ if and only if there exists a finite subcollection of $\{U_\alpha\}_{\alpha\in I}$ that covers $[0,x]$, that is, $$[0,x]\subset\bigcup\limits_{i=1}^n U_{\alpha_i}.$$
Prove that if $x\notin P$ then there exists a neighborhood $O$ of $x$ such that $O\cap P=\varnothing$.

This problem proceeds after this one posted here
My attempt at the proof is,
Suppose $x\not\in P$. Then $x\in C(P)$ and since $P$ is open, $C(P)$ must be closed. Because $C(P)$ is closed, it contains the boundary of $P$. Thus, $x\in\text{Bdry}(P))$ or $x\in\text{Int}(C(P))$. If $x\in\text{Int}(C(P))$ then there exists an open set $O\subset\text{Int}(C(P))$ such that $O\cap P=\varnothing.$
I'm having trouble showing that the part with $x\in\text{Bdry}(P)$. I know that since its already shown earlier that $P$ is open and this proof will show that $P$ is closed that ultimately $\text{Bdry}(P)=\varnothing$, which means $x$ can't be in $\text{Bdry}(P)$.
Is there a way to avoid this process and show from the beginning that $x$ must lie in $\text{Int}(C(P))$? I'm also thinking of using the definition of $P$ to come up with something and so far I was able to show that there exists a $p\in[0,x]$ such that $p\in\bigcap\limits_{i=1}^n C(U_{\alpha_i})$, since $x\notin P$, yet I don't know where to go from there.
Thanks for any hints to this problem.
 A: Let's assume otherwise for a contradiction. We'll prove $x\in P$. Of course, there's an element $U_x$ in the cover containing $x$. $U_x$ must contain a point $p\in P$. The set $[0,p]$ admits a finite subcovering by the given open cover and together with $U_x$, we have an finite subcovering of $[0,x]$. In other words, $x\in P$ which is a contradiction.
The proof is straightforward once you use all the hypothesis (e.g., the definition of $P$ which you seem to have forgotten to use in your proof). 
Exercise 1: We've proved that $P$ is closed. Can you use a similar argument to prove that $P$ is open in $[0,1]$?
Exercise 2: Use the connectedness of $[0,1]$ to deduce the compactness of $[0,1]$.
Exercise 3: In order to complete the proof of the compactness of $[0,1]$, we need to establish the connectedness of $[0,1]$. Let $\{U,V\}$ constitute a non-trivial separation of $[0,1]$. Let's assume without loss of generality that $0\in U$. Prove that the least upper bound of $U$ is an element of $U\cap V$; a contradiction. Therefore, $[0,1]$ is connected.
I hope this helps!
