$[0,1]$ and the Cantor set are compact, by Heine-Borel and by closedness inside $[0,1]$; and, assuming the Axiom of Choice, so is $[0,1]^{\mathbb{N}}$, by Tychonoff

Let $\mathbb{R}$ carry its usual topology (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y) = |x-y|$ is a metric, its open balls are the intervals $(x-r, x+r)$, and it is unbounded, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not), let $[0,1]$ (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length) and the Cantor set $C$ (The Cantor middle-thirds set as the intersection of the sets $C_n$ obtained by removing open middle thirds) carry the subspace topology (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace), and let

$Q \;:=\; [0,1]^{\mathbb{N}} \;=\; \prod_{n \in \mathbb{N}} [0,1]$

carry the product topology (The product set $\prod_{i \in I} X_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space). Then, with compactness as in Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right:

1. $[0,1]$ is compact.
2. $C$ is compact, being closed inside $[0,1]$.
3. $Q$ is compact, assuming the Axiom of Choice.

Each of the three uses a different tool, and that is the point of putting them together: Heine-Borel for a closed bounded subset of the line, the closed-subspace theorem for a closed subset of something already known to be compact, and Tychonoff for a product over an infinite index set.

Claim 2 does not need Heine-Borel a second time. The Cantor set is closed and bounded, so [L1] would give its compactness directly; the route through $[0,1]$ is taken because it uses only that $C$ is closed in a space already known to be compact, which is the argument that generalises to spaces with no metric.

Claim 3 is where the cost appears. Claims 1 and 2 are theorems of ZF, the bisection proof of Heine-Borel selecting nothing; claim 3 rests on Tychonoff's theorem: an arbitrary product of compact spaces is compact in the product topology, assuming the Axiom of Choice and therefore on the Axiom of Choice. For a product over a finite index set no choice is needed (A product of finitely many compact spaces is compact in the product topology), so the cost is attached to the infinite index set and not to the factors.

$Q$ is metrizable, and that is a separate fact. Compactness of $Q$ is proved here from the product structure alone and uses no metric on $Q$; whether a metric inducing the product topology exists is a separate question, which nothing among this page's declared prerequisites answers and which no claim above needs.