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 $X.$
My question is how to prove formally that each infinite subset of $X = \mathbb{Z}_{+} \times Y$ has a limit point?
I can see it but I don't know if that's enough for a formal justification:
$U_1=1\times Y=\{(1,y_1),(1,y_2):y_1,y_2\in Y\},U_2= \dots $
Thus $(1,y_1)\in U_1$ is a limit point of $U_1$ because each open set that contains $(1,y_1)$ also contains $(1,y_2)$. Similar with the others $U_i$.
It's actually true (as it says in the text of the counterexample) that every nonempty subset has a limit point, regardless if it's infinite. Let $S$ be a nonempty subset of $\mathbb Z_+\times Y$ and let $(n,y)\in S.$ Then $(n,\bar y)$ is a limit point of $S,$ where $\bar y$ is the other element of $Y.$ The neighborhoods of $(n,\bar y)$ are sets of the form $A\times Y$ where $n\in A$ and thus $(n,y)$ is an element of $S$ that is in every neighborhood of $(n,\bar y)$.