Trying to understand a set inclusion I got stuck while reading a proof from Michael Field's Essential Real Analysis. Here is the difficult part of his text:
"Choose $\varepsilon>0$. Since $(X_n)$ is Cauchy, there exists $N_1\in\mathbb{N}$ such that $h(X_n,X_m)<\varepsilon$, for all $n,m\geq N_1$. Since $X_m\subset X_n(\varepsilon)$, for all $n,m\geq N_1$ and $\Lambda=\bigcap_{n\geq p}\overline{\bigcup_{m\geq n}X_m}$, all $p\geq 1$, we certainly have $\Lambda\subset X_n(\varepsilon)$, $n\geq N_1$."
What I don't get is how "... we certainly have $\Lambda\subset X_n(\varepsilon)$, $n\geq N_1$."
$h(X_n,X_m)<\varepsilon\Rightarrow X_m\subset X_n(\varepsilon)$ follows from a lemma and $\Lambda$ is defined earlier as $\Lambda=\bigcap_{n\geq 1}\overline{\bigcup_{m\geq n}X_m}$. I can't catch what Field means with switching "$1$" to "$p$" and with "all $p\geq 1$".
$h$ is a metric so $h(X_n,X_m)=h(X_m,X_n)$ and $X_n\subset X_m(\varepsilon)$ also. The sets $X_n$ that constitute the sequence $(X_n)$ are compact subsets of $\mathbb{R}^n$, thus they are closed and bounded. $X_n(\varepsilon)$ is an open neighbourhood of $X_n$. That means all the points within $\varepsilon$ of $X_n$.
I've tried to figure this out for the whole day and I'd appreciate help a lot.
EDIT: I added a few points to make the question clearer.
 A: Let $Y_n=\overline{\bigcup_{m\geq n}X_m}$.  Note that if $n\leq n'$ then $Y_n\supseteq Y_{n'}$, since $Y_{n'}$ has fewer terms in the union.  So when we define $$\Lambda=\bigcap_{n\geq 1}Y_n,$$ that is actually the intersection of a nested sequence $Y_1\supseteq Y_2\supseteq Y_3\supseteq\dots$.  This means that if you start the sequence at $Y_p$ rather than $Y_1$, the intersection is still the same, since $Y_p$ is already contained in all of $Y_1,\dots,Y_{p-1}$.  That is, $\Lambda$ is also equal to $$\bigcap_{n\geq p}Y_n$$ for any $p\geq 1$.

Actually, though, this step is unnecessary.  In order to conclude that $\Lambda\subseteq X_n(\epsilon)$ for $n\geq N_1$, you can just note that $Y_n\subseteq X_n(\epsilon)$ since $Y_n$ is the closure of a union of sets, all of which are contained in $X_n(\epsilon)$. (I assume here that however $X_n(\epsilon)$ is defined, it is a closed set.  If it isn't, then the claim that $\Lambda\subseteq X_n(\epsilon)$ may be wrong and needs to be replaced with something like $\Lambda\subseteq X_n(2\epsilon)$.)  Since $Y_n$ is one of the terms in the intersection defining $\Lambda$, this immediately implies that $\Lambda\subseteq X_n(\epsilon)$.
