Understanding proof that $X$ is compact if it is a metric space in which every infinite subset has a limit point 
Let $X$ be a metric space in which every infinite subset has a limit point. Prove that $X$ is compact.
Proof: $X$ has a countable base. It follows that every open cover of $X$ has a countable subcover $\{G_n\}$, $n = 1, 2, 3, ...$. If no finite subcollection of $\{G_n\}$ covers $X$, then the complement $F_n$ of $G_1 \cup ... \cup G_n$ is nonempty for each $n$, but $\bigcap F_n$ is empty. If $E$ is a set which contains a point from each $F_n$, consider a limit point $p$ of $E$. All but finitely many points of $E$ are in $F_n$, so $p$ is a limit point of each $F_n$. But each $F_n$ is closed, so $p$ is in every $F_n$, which is a contradiction. So $X$ is compact.

The part I don't understand is why $p$ is a limit point of each $F_n$ simply because it is a limit point of $E$ and infinitely many point of $E$ are in $F_n$. i.e. I don't see the justification of why $p$ is in the closure of each $F_n$.
 A: Not "infinitely-many" points of $E$ (though this is true); rather, "all but finitely-many." This is an important distinction.
Put another way, it means that there are only finitely-many points of $E$ that don't lie in $F_n.$ Note, then, that since $E\setminus F_n$ is finite, then there is some $R>0$ such that $$d(x,p)<R\implies x\notin E\setminus F_n.$$ Hence, if $x\in E$ and $d(x,p)<R,$ then $x\in F_n.$
Now, take any $0<r<R.$ Since $p$ is a limit point of $E,$ then there exists $x\in E$ such that $d(x,p)<r,$ and so there exists $x\in F_n$ such that $d(x,p)<r.$ Thus, every open ball about $p$ intersects $F_n,$ and so....

Added: Consider the family $F_n=\left[-\frac1n,\frac1n\right]$ of closed intervals in $\Bbb R,$ and the set $$E=\left\{\frac1n:n\in\Bbb Z,n\ge1\right\}\cup\left\{2-\frac1n:n\in\Bbb Z,n\ge 1\right\}.$$ Readily, infinitely-many points of $E$ are in each $F_n,$ but while $2$ is a limit point of $E,$ it is not a limit point of any of the $F_n$s, which is possible because for a given $F_n,$ there are also infinitely-many points of $E$ that do not lie in $F_n.$
