$\mathbb{R}$ covered by its closed singletons: every restriction of the indicator of $\{0\}$ is continuous and the map is not, so the closed pasting lemma needs finiteness

Refuted: that continuity may be checked on an arbitrary closed cover. Claim 3 of Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous allows only finitely many closed pieces, and the restriction is not removable.

Witness. Give $\mathbb{R}$ its usual topology (Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not, 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) and let

$\mathcal{F} := \{\, \{t\} : t \in \mathbb{R} \,\}$

be the family of its singletons, a cover of $\mathbb{R}$ by closed sets. Let $f : \mathbb{R} \to \mathbb{R}$ be the indicator of $\{0\}$, that is $f(0) = 1$ and $f(t) = 0$ for $t \ne 0$. Then every restriction $f|_{\{t\}}$ is continuous for 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 $f$ is not continuous (Continuity of a map of topological spaces at a point and globally).

• Why an infinite closed cover is useless and an infinite open cover is not. The proof of claim 3 of Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous writes $f^{-1}[F]$ as a union of sets closed in $\mathbb{R}$ and concludes that it is closed; only finite unions of closed sets are closed (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison), and here $f^{-1}[\mathbb{R} \setminus V]$ is the union of the uncountably many closed sets $\{t\}$, $t \ne 0$, which is $\mathbb{R} \setminus \{0\}$, not closed. The open-cover version has no such restriction because arbitrary unions of open sets are open.

• The singleton cover trivialises every function. For any spaces $X$ and $Y$ and any $f : X \to Y$, the restriction of $f$ to a one-point subspace is continuous, so the singleton cover certifies nothing whatever. The witness is therefore the sharpest form of the failure rather than a delicate example, and the map $f$ could be replaced by any discontinuous function.

• A two-piece closed cover of $\mathbb{R}$ would have detected the discontinuity. For instance $(-\infty,0]$ and $[0,\infty)$ are closed and cover $\mathbb{R}$, and $f$ restricted to $(-\infty,0]$ is already discontinuous at $0$ by the argument of step 1.4 carried out inside that subspace.