If $X\neq\varnothing$ and $\tau=\{\varnothing, X\}$, then any subset of $X$ is compact.

Question

If $X\neq\varnothing$ and $\tau=\{\varnothing, X\}$, then any subset of $X$ is compact.

Disproof by counterxample? Not true. Let $X = \mathbb{R}$ with the usual topology and $A = (-\infty,0)$. Clearly $A$ is not compact.

EDIT: Thank you to those below who provided their criticism and feedback.

Here is my new attempt

Let $A \subset X$ and $\{G_{\alpha}\}_{\alpha \in a}$ be an open cover of A.

Since $\emptyset$, $X$ are the only open sets in $\tau$, $G_{\alpha} = X \cup \emptyset = X$

As a result, there is a finite subcover of $A$, hence $A$ is compact

Therefore, any subset of $X$ is compact.

Answer 1

Hint:

Can there be an open cover of $A\subset X$ that is not finite?

Answer 2

Compactness is something relative to the topology.

You are given a topology on $X$, namely $\tau=\{\varnothing,X\}$. And the claim is that every subset of $X$ is compact in that topology.

So your counterexample is not a counterexample, since it doesn't uses the given topology.

Answer 3

It depends on definition. It may be:

1. $X$ is compact if each of its open covers has a finite subcover.

or

2. $X$ is compact if it is Hausdorff ($T_2$) and each of its open covers has a finite subcover.

There is only one nonempty open set, hence $X$ is compact in the sense of 1.

But even if $X=\{0,1\}$ with the topology in question, it is not Hausdorff, hence is not compact in the sense of 2.