I'm reading 'Measure Theory' from Bogachev and have a problem understanding
one of its proofs. First a definition: A set of subsets $K$ of a set is a
compact class if for any sequence $\{ K\_i \}\_{i \in {N}\_0}$ in $K$
with $\bigcap\_{i = 1}^{\infty} K\_i = \emptyset$ there exists a $N \in
{N}\_0$ such that $\bigcap\_{i = 1}^N K\_i = \emptyset$. As an example
they say that a set of compact sets is a compact class. To prove this they
proceed by contradiction so let $\{ K\_i \}\_{i \in {N}\_0}$ be a family
of sets in $K$ with $\bigcap\_{i = 1}^{\infty} K\_i$ and let $\forall n \in
{N}\_0$ $\bigcap\_{i = 1}^n K\_i \neq \emptyset$ then we can construct a
$\{ x\_i \}\_{i \in {N}\_0}$ with $\forall n \in {N}\_0$ $x\_n \in
\bigcap^n\_{i = 1} K\_i$, no element of $\{ x\_i |i \in {N}\_0 \}$ can be
repeated infinitely. So far I follow the proof, however then the book says
that because $K\_1$ is compact there exists a $x$ such that every neighborhood
of $x$ intersects infinitely elements of $\{ x\_i \}\_{i \in \infty}$ which does
not seem obvious to me. The only thing I can come up with is the following:
$\{ x\_i | i \in {N}\_0 | \}$ is infinite, $K\_1$ is compact thus
sequential compact so $\{ x\_i | i \in {N}\_0 \}$ has a limit
point $x$, but htis only means that every neighborhood contains a point other
then $x$ in $\{ x\_i |i \in {N}\_0 \}$ which is different from
containing infinite elements. Any help with this (or an alternative proof for
this example) ?

Thanks a lot in advance

Marc