Rational comb with a deleted attachment point ($C=(\mathbb{Q}\times [0,1])\cup (\mathbb{R}\times \{1\})$)

The displayed subspace of the Euclidean plane is path-connected: each vertical rational tooth joins the horizontal line at height one. Delete $(r,1)$ for rational $r$. The residual tooth $V=\{r\}\times [0,1)$ cannot be joined by a path to the top line. Before a path first reaches height one, its first coordinate is a continuous rational-valued function on an interval and must therefore be constant; it would reach the deleted point.

The punctured space remains connected. Its portions $L,R$ on either side of $x=r$ are path-connected, and $V\subseteq \overline{L}\cap \overline{R}$ by the density of the rational numbers. Adjoining limit points to a connected set preserves connectedness: a separation would place the connected subset on one side and prevent the other side from containing its limit points. Hence $L\cup V$ and $R\cup V$ are connected and intersect in $V$. Their union is connected. This example distinguishes connectedness from path-connected using the loss of a single attachment point.