Proving a space compact takes an argument about every cover. Denying it takes one cover and a reason it cannot be thinned, which is much less work and settles the matter.

To claim every key on the ring is spare you must try them all. To deny it, one door that only one key opens is enough.

Compactness is a claim about every cover, so it has to be argued. Its denial is a claim about one cover, so a single example settles it. Two failures are worth carrying: the real line, and an infinite set with the discrete topology. The closed interval $[0,1]$ is compact; that is stated here and not proved, because its proof needs the least upper bound property of the real numbers, which belongs to the analysis course and cannot be borrowed into this one.

• The intervals $(-n,n)$ are open and every real number lies in one of them, so the union above is a genuine open cover.
• Finitely many of those intervals have a largest, and their union is that largest one, which leaves out its own right-hand endpoint.
• In an infinite discrete space the single points form an open cover, and dropping even one of them leaves a point uncovered.
• $[0,1]$ is compact. The result is used here without proof and the reason is stated plainly rather than hidden.

Every real sits in some $(-n,n)$, so these intervals cover $\mathbb{R}$. No finite number of them does: the widest chosen would have a right-hand end and the line does not. One cover that cannot be thinned is the whole proof that $\mathbb{R}$ is not compact.

Cover the line by the intervals running from minus to plus each whole number. They are open, and every real number is smaller in size than some whole number, so they cover. Now take any finite handful of them. Finitely many whole numbers have a largest, and since the intervals grow as the number grows, the union of the handful is simply the largest interval in it. That interval leaves out the very number that names it. So no finite subfamily covers the line, and one cover that cannot be thinned is all it takes.

Each interval is open, and every real number lies in one of them, so this is an open cover.

The intervals grow with the number, so a finite handful of them has a largest.

Being nested, the union of finitely many is the largest of them, where $N$ is the largest index.

The number $N$ is left out, so no finite subfamily covers, and the line is not compact.

The one cover that settles the real line.

Nested sets: the union is simply the biggest.

A cover in which nothing is spare.

Used here, proved elsewhere.