Past exam of the metric spaces and complex analysis course of the University of Oxford / 2015 / 1 / c

The sequence $n\in [0,\infty )$ is unbounded. Every compact metric space is bounded, by part (b), so this subspace is not compact.

Take an irrational number $a\in (0,1)$. The density of the rational numbers provides a sequence in $\mathbb{Q}\cap [0,1]$ converging to $a$, which is outside that subset. Thus the subset is not closed in $\mathbb{R}$. By the Heine-Borel theorem, it is not compact, despite being bounded.

The unit circle is the inverse image of the closed set $\{1\}$ under the continuous map $(x,y)\mapsto x^2+y^2$, so it is closed. It is also bounded, since every point has Euclidean norm one. The Heine-Borel theorem gives compactness.

The open unit disk is bounded but not closed: $(1-1/n,0)$ lies in it for $n\ge 2$ and converges to the omitted point $(1,0)$. By the Heine-Borel theorem, it is not compact.

The points $(n,0)$ belong to the indicated subset for every positive integer $n$. They form an unbounded sequence, so this metric space is not bounded. Therefore it is not compact. Shrinking the vertical width does not bound its horizontal extent.