Error in Engelking? I've been trying to do some exercises in Engelking's General Topology text, and there's one that's causing me problems.  I hope that someone here can clarify this for me.
The exercise is (slightly paraphrased):

1.5.N Let $\mathcal{U}$ be a point-finite open cover of a topological space $X$.  Show that the set $$L = \{ x \in X : x\text{ has a neighbourhood }V\text{ that meets only finitely many members of }\mathcal{U} \}$$ can be represented as the intersection of countably many dense open subsets of $X$.

There is a hint given, which reads (again slightly paraphrased):

Hint. For every $x \in X$ let $n(x) = | \{ U \in \mathcal{U} : x \in U \} |$ and consider the open sets $$G_i = \{ x \in X : n(x) > i\text{, or }x\text{ has a neighbourhood }V\text{ such that }n(y) \leq i\text{ for each }y \in V \}.$$

To me it seems like the $G_i$ should be the open dense subsets of $X$ that we are looking for in the exercise (and it is pretty easy to show that they are dense).  However I have a problem showing that $L = \bigcap_i G_i$.
In fact, this may not even be true!  Consider $\mathcal{U} = \{ ( - \infty , 0 ) , ( -1 , 1 ) , ( 0 , + \infty ) \}$.  Clearly this is a (point-)finite open cover of the real line.  It is easy to show that $L = \mathbb{R}$ (since we have a finite open cover), and $$n(x) = \begin{cases}
1, &\text{if }x \leq -1 \\
2, &\text{if }-1 < x < 0 \\
1, &\text{if }x = 0 \\
2, &\text{if }0 < x < 1 \\
1, &\text{if }x \geq 1.
\end{cases}$$
From here we can show that


*

*$G_1 = \mathbb{R} \setminus \{ -1 , 0 , 1 \}$ and

*$G_i = \mathbb{R}$ for all $i > 1$


and so $L = \mathbb{R} \neq \mathbb{R} \setminus \{ -1 , 0 , 1 \} = \bigcap_i G_i$.
Of course, in this example the set $L$ can trivially be written as the intersection of countably many dense open subsets of $\mathbb{R}$: $L = \bigcap_i \mathbb{R}$.  So I don't have an actual counterexample.
From my interpretation of the hint, it appears that the best I can do is show that $L$ has a subset which can be expressed as the intersection of countably many dense open subsets of $X$.
So my question is whether Engelking's exercise is correct (and if so how would you prove it).  Otherwise I'd like an actual counterexample.
Thank-you
 A: What you need is the $\limsup G_i$ in a set theoretical sense. The open dense sets to be intersected are $H_n = \bigcup_{i>n}G_i$.
In other words, $L$ is the set of points that are eventually in $G_i$.
A: Another method is to mainly ignore the hint!
First note that $L = \bigcup \{ U \subseteq X : U\text{ is open and meets only finitely many elements of }\mathcal{U} \}$, and so $L$ is actually open itself.
We now define a decreasing family $\{ V_n : n \in \mathbb{N} \}$ of open subsets of $X$ such that


*

*$L \cup V_n$ is dense for all $n \in \mathbb{N}$; and

*$\bigcap_{n \in \mathbb{N}} V_n = \varnothing$.


(Then $L = \bigcap_{n \in \mathbb{N}} ( L \cup V_n )$, as desired.): 
For this we define $$V_n = \bigcup \{ U_1 \cap \cdots \cap U_n : U_1 , \ldots , U_n \in \mathcal{U}\text{ are distinct} \},$$ that is $V_n$ is the union of all intersections of $n$ distinct sets from $\mathcal{U}$.
Clearly each $V_n$ is open, and $V_{n+1} \subseteq V_n$ for all $n \in \mathbb{N}$.  If $x \in \bigcap_{n \in \mathbb{N}} V_n$, then it must be that $x$ belongs to infinitely many elements of $\mathcal{U}$, contradicting that $\mathcal{U}$ is point-finite.
Finally, given $n \in \mathbb{N}$, to see that $L \cup V_n$ is dense let $U \subseteq X$ be an arbitrary nonempty open set.  There are two cases:


*

*$U \cap L \neq \varnothing$; or

*$U \cap L = \varnothing$.


In the former case there is nothing to do.  In the latter case it follows that every nonempty open subset of $U$ meets infinitely many sets in $\mathcal{U}$, and from this we can inductvely pick distinct $U_1 , \ldots , U_n \in \mathcal{U}$ such that $U \cap ( U_1 \cap \cdots \cap U_n )$ is nonempty.
