Why is this set compact?

Why is this set compact?

Let $x_n$ converge in a metric space to $x$, and $F_n=\{x, x_n,x_{n+1},
x_{n+1}, ...\}$.
Why is $F_n$ compact? I was going to invoke the following theorem "A
subset $A$ of a metric space is compact iff. every infinite subset of $A$
has an accumulation point in $A$". But it doesn't even seem to me now that
the sequence limit $x$ must be contained in every infinite subset of $A$:
for example $\{x_n,x_{n+1}, x_{n+1}, ...\}$ is an infinite subset of
$F_n$, but without knowing if this subset is closed or not, how do we know
whether $x$ is in this set?