Example 6 in Munkres' book

In Topology, the second edition by Munkres, in section 22, on page 143 he says the following:

"Example 6.$\quad$ Let $X$ be the subspace of $\mathbb{R}^2$ that is the union of the line segments $[0,1]\times \{n\}$, for $n\in \mathbb{Z}_+$, and let $Z$ be the subspace of $\mathbb{R}^2$ consisting of all points of the form $x\times (x/n)$ for $x\in[0,1]$ and $n\in \mathbb{Z}_+$. Then $X$ is the union of countably many disjoint line segments, and $Z$ is the union of countably many line segments having an end point in common.

Define a map $g:X\rightarrow Z$ by the equation $g(x\times n)=x\times (x/n)$; then $g$ is surjective and continuous. The quotient space $X^*$ whose elements are the sets $g^{-1}(\{z\})$ is simply the space obtained from X by identifying the subset $\{0\}\times \mathbb{Z}_+$ to a point. Then map $g$ induces a bijective continuous map $f:X^*\rightarrow Z$. But $f$ is not homeomorphism.

To verify this fact, it suffices to show that $g$ is not a quotient map. Consider the sequence of points $x_n=(1/n)\times n$ of $X$. The set $A=\{x_n\}$ is a closed subset of $X$ because it has no limit points. Also, it is saturated with respect to $g$. On the other hand, the set $g(A)$ is not closed in $Z$, for it consists of the points $z_n=(1/n)\times (1/n^2)$; this set has the origin as a limit point."

Why is subset $A=\{x_n\}$ closed, while it has no limit points? As far as I know, a subset of a topology space is closed if and only if it contains all its limit points. I think the author has a mistake.

Can someone help me? Thanks.