TITLE: Uncountable Set with Cofinite Topology
QUESTION [6 upvotes]: In this proofwiki page, it is written without any justification that $\cap \mathcal{B}_x = \{x\}$ where $\mathcal{B}_x$ is some countable basis of a point $x$ in the uncountable set $S$ endowed with the cofinite topology. Why is this true? Indeed, I just proved that it cannot be true in an uncountable space with the cofinite topology. Is this proof flawed? This steps seems crucial in obtaining a contradiction. 
What I am trying to do is prove that if $X$ is an uncountable set with the cofinite topology, then it cannot be first countable. I read this, but I couldn't see how to use Brian Scott's hints. Here is what he wrote:

To prove that $X$ is not first countable, you must show that some point of $X$ does not have a countable local base. All points of $X$ ‘look alike’ in the cofinite topology, so it doesn’t matter what point we pick, so let $x\in X$ be any point. Suppose that $\mathscr{B}=\{B_n:n\in\Bbb N\}$ is a countable local base of open sets at the point $x$, meaning that if $U$ is any open nbhd of $x$, then $x\in B_n\subseteq U$ for some $n\in\Bbb N$. For each $n\in\Bbb N$ let $F_n=X\setminus B_n$: $B_n$ is open, so by definition $F_n$ is finite. Let $F=\{x\}\cup\bigcup_{n\in\Bbb N}F_n$; $F$ is the union of countably many finite sets, so $F$ is countable. $X$ is uncountable, so there is some $y\in X\setminus F$. Let $U=X\setminus\{y\}$.

Is $U$ an open nbhd of $x$?  
Is there any $n\in\Bbb N$ such that $x\in B_n\subseteq U$?


I would answer the first question affirmatively; and to answer the question, which is presumably "no", I would suppose that there is such an $n \in \Bbb{N}$. But I am having trouble identifying the contradiction. 
Before I even sought help from google, I tried the following (and several variations on it). Let $x \in X$, and let $y \in X - \{x\}$. Then $X-\{y\}$ is open and therefore there is an open set $U_y$ such that $x \in U_y \subseteq X - \{y\}$. Thus we have a mapping $y \mapsto U_y$. I tried to show this mapping is injective, but I couldn't. If I had succeeded, then I believe this would have shown that if there were any basis at $x$, it would have to be uncountable, which means that $X$ is not first countable. But, as you may have noticed, I abjectly failed in this task. 
Is there a way to get any of these to proves to work, especially the one I proposed?

REPLY [4 votes]: The proof just uses the following general fact:

If $X$ is $T_1$ (or equivalently all sets of the form $\{x\}$ are closed) and $x \in X$ and $\mathcal{B}$ is any local base at $x$, then $\bigcap_{B \in \mathcal{B}} B = \{x\}$.

Proof: Of course $x$ is that intersection, so that's a trivial inclusion. Suppose that $p \in \bigcap_{B \in \mathcal{B}} B$ and that $p \neq x$. Then $\{p\}$ is closed and so $x \in X\setminus\{p\}$ which is open so (as we have a local base) for some $B_p \in \mathcal{B}$ we have $x \in B_p \subseteq X\setminus\{p\}$, so $p \notin B_p$, which contradicts $p \in \bigcap_{B \in \mathcal{B}} B$, so $p$ must equal $x$, which proves the other inclusion. 
The cocountable topology is $T_1$ as one point sets are countable so closed.