I think that there should be a comma after Lindelöf. Other than that, all I can do is paraphrase the statement, hoping that this helps.
Suppose $X$ is not Lindelöf. In particular, $X$ is not compact. Since $X$ is Tychonoff, $X$ has a Stone-Cech compactification $\beta X$.
As usual, we consider $X$ to be a subset of $\beta X$ and can now form the Stone-Cech remainder $\beta X\setminus X$. The remainder is nonempty since $X$ is not compact.
Now the claim is the following: There is a compact set $C\subseteq\beta X\setminus X$ with the following property:
For all $G_\delta$-sets $A\subseteq\beta X$ with $C\subseteq A$, $A\cap X\not=\emptyset$.
A $G_\delta$-set is an intersection of countably many open sets.
I hope this clarifies something.
You can get this compact set as follows: Let $\mathcal U$ be an open cover of $X$ without a countable subcover. This is possible since $X$ is not Lindelöf. We can assume that the $U\in\mathcal U$ are actually open subsets of $\beta X$ so that no countable subcollection of $\mathcal U$ covers $X$.
$\mathcal U$ is not a cover of $\beta X$ since in this case, by compactness of $\beta X$, finitely many elements of $\mathcal U$ would already cover $\beta X$ and in particular $X$.
It follows that the compact set $C=\beta X\setminus\bigcup\mathcal U$ is nonempty.
Let $A$ be a $G_\delta$ subset of $\beta X$ with $C\subseteq A$.
Since $A$ is $G_\delta$, $\beta X\setminus A$ is the union of a countable family $\mathcal B$ of closed sets. Each $B\in\mathcal B$ is compact and disjoint from $C$. Since $C=\beta X\setminus\bigcup\mathcal U$, each $B\in\mathcal B$ is covered by $\mathcal U$ and hence by finitely many elements of $\mathcal U$.
It follows that $\bigcup\mathcal
B$ is covered by countably many elements of $\mathcal U$.
But $X$ is not covered by countably many elements of $U$. It follows that there is $x\in X\setminus\bigcup\mathcal B=A\cap X$.
So $A$ meets $X$. This finishes the proof.