In the bounded real-valued functions on $\mathbb{N}$ with the supremum metric, the closed unit ball is closed and bounded and is not compact: the indicator functions of the singletons are pairwise at distance $1$

Refuted claim: in every metric space a closed and bounded subset is compact (FALSE: a closed and bounded subset of a metric space is compact).

The witness is the space $\mathcal{B}(\mathbb{N})$ of bounded functions $\mathbb{N} \to \mathbb{R}$ with the supremum metric $d_\infty(f,g) = \sup\{\,|f(s)-g(s)| : s \in \mathbb{N}\,\}$ (The supremum metric $d_\infty(f,g) = \sup_x |f(x) - g(x)|$ is a metric on the bounded real-valued functions on a nonempty set), together with the closed unit ball

$K \;:=\; \{\, f \in \mathcal{B}(\mathbb{N}) : d_\infty(f, \mathbf{0}) \le 1 \,\}$

about the zero function $\mathbf{0}$. The set $K$ is closed in $\mathcal{B}(\mathbb{N})$ (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement) and bounded (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space), and it is not compact (Open cover, subcover, compact metric space, and compact subset of a metric space): the indicator functions $e_n$ of the singletons $\{n\}$ lie in $K$ and satisfy $d_\infty(e_m,e_n) = 1$ whenever $m \ne n$, so no finite $1/2$-net for $K$ can exist.

This is the same failure as in FALSE: a closed and bounded subset of a metric space is compact, in the space where it matters analytically: closed bounded sets of a function space are routinely not compact.

What goes wrong is room, not size. The ball $K$ has diameter $2$, so it is small in the metric sense; but it contains infinitely many points that are pairwise at distance $1$, so no finite family of small balls reaches all of them. That is exactly the failure of total boundedness, and it is one of the two conditions that, by For a metric space, compact, countably compact, limit point compact, sequentially compact, and complete together with totally bounded are all equivalent, given countable choice and dependent choice, are together equivalent to compactness once the Axiom of Countable Choice and the Axiom of Dependent Choice are assumed.

What the argument does and does not settle. It shows that $K$ fails total boundedness, and that alone rules out compactness by [L3]. Nothing above is claimed about whether $K$ is complete; the point of the witness is that closedness and boundedness, the two conditions that suffice in $\mathbb{R}^n$ (Heine-Borel in $\mathbb{R}^n$: with the Euclidean metric a subset of $\mathbb{R}^n$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line), do not suffice here.