A $T_{3 \frac{1}{2}}$ Lindelöf space $X$ is normally placed in $\beta X$

Question

A subset $X \subset Y$ is normally placed in $Y$ if whenever $X \subset U$ for $U$ open in $Y$ there are $F_n$ closed in $Y$ such that $X \subset \bigcup_{n \in \mathbb{N}}F_n \subset U$.

Let $X$ be a $T_{3 \frac{1}{2}}$ space, and assume it is Lindelöf. Why is it normally placed in its Stone-Čech compactification $\beta X$?

My only lead so far was to write $U$, open in $\beta X$, as a union of basic open sets of intervals, and find closed intervals inside them - but this ignores the fact that $X$ may be open itself.

Hints, please?

Answer 1

Each $x \in X$ has open neighborhood $V_x$ in $\beta X$ such that $\overline{V}_x \subset U$.

$\{ V_x \cap X \}_{x \in X}$ is an open cover of $X$, hence it has a countable subcover $\{ V_{x_n} \cap X \}_{n \in \mathbb N}$. Now take $F_n = \overline{V}_{x_n}$.

Note that a Lindelöf $X$ is normally placed in any regular $Y \supset X$.