Let $(X,\tau)$ be a topological space. If $X$ has the Bolzano-Weierstrass property then $X$ is compact.

Question

Where the indiscrete topology was used (in the counterexample)?

Let $(X,\tau)$ be a topological space. If $X$ has the Bolzano-Weierstrass property then $X$ is compact.

Counterexample

Let $Y$ consist of two points; give $Y$ the topology consisting of $Y$ and the empty set. Then the space $X = \mathbb{Z}_{+} \times Y$ is limit point compact, for every nonempty subset of $X$ has a limit point. It is not compact, for the covering of $X$ by the open set $U_n = \{n\} \times Y$ has not finite collection covering $X.$

Answer 1

It is used in the highlighted portion.

give $Y$ the topology consisting of $Y$ and the empty set. Then the space $X = \mathbb{Z}_{+} \times Y$ is limit point compact, for every nonempty subset of $X$ has a limit point.

That is true only because in Y, every point is a limit point because it has the indiscrete topology.

Let the points of Y be named $y_1$ and $y_2$. For any point $z=(n, y_1)\in X$, any neighborhood of $z$ must also contain the point $(n, y_2)$ (and vice versa). So every point of $X$ is a limit point. Note that the author has used the product space notation for X, I.e., the author is using the topology on $X$ which is a product of the discrete topology of $\mathbb{Z}_+$ and whatever topology is put on $Y$.

As to what happens if Y did not have the indiscrete topology... there are only 3 other possible topologies - the discrete topology and the two Sierpiński topologies. In all those cases, you can find a singleton set that is open in X and it is no longer true to say “every nonempty subset of $X$ has a limit point”.

Answer 2

Points in $X$ look like $(n,y)$ with $n \in \mathbb Z$ and $y \in Y = \{y_1, y_2\}$.

Let $A$ be a nonempty subset of $X$ and let $x \in A$. If $U$ is an open set containing $x$, then it must be of the form $B \times Y$ with $B \subset X$ due to the topology of $Y$.

If $x = (n,y_1)$ for $x \in \mathbb Z$, then $x$ is a limit point since $U - \{x\}$ contains the point $(n,y_2)$ and is therefore nonempty. Hence every point in $X$ is a limit point.

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Edit: The topology on $X$ is the natural product topology of the discrete topology on $\mathbb Z_+$ and the indiscrete topology on $Y$. That is, a set $U = G \times Z$ is in the topology of $X$ if $G$ is in the topology of $\mathbb Z_+$ (any subset) and $Z$ is in the topology of $Y$ (i.e. $\emptyset$ or $Y$).