How can I prove this? Why can't we use the proof for [x,y] and Supremum here? I suppose because it is not bounded, and we don't know that we have a Supremum/Infimum at all (which we don't). On the other side I don't see how to use the counterexample for $(x,y)$ or $(x,y]$ here.

Definitions:

1. A topological space $X$ is called compact if each open coverage of $X$ has a finite partial coverage, i.e. $I$ is a set and $U\_i \subseteq X$ is open
$(i \in I)$ with $\bigcup\_{i\in I} U\_i = X$, then a finite subset $J \subseteq I$ exists with $\bigcup\_{j\in J} U\_j = X$.
2. A subset $A$ of a topological space $X$ is called compact, if $A$ is compact in the track topology, i.e. if for a set $I$ and open sets $U\_i (i ∈ I)$ with $\bigcup\_{i\in I} U\_i \supseteq A$ there is always a finite subset $J \subseteq I$, so that $\bigcup\_{j\in J} U\_j \supseteq A$ applies

Edit: I would like to prove this using this definitions. Means w/o H-B.