 Solution

Write $C^{\prime }=C\setminus \{(1,1)\}$ and $V=\{1\}\times [0,1)$. Let $L$ and $R$ be the portions of $C^{\prime }$ with first coordinate respectively less than and greater than one. Each is path-connected: move along a rational tooth to its corresponding open half of the top line, then horizontally there. By the density of the rational numbers, every point of $V$ is a limit point of both $L$ and $R$.

Use the elementary fact that adjoining any collection of limit points to a connected set preserves connectedness. Indeed, in a proposed separation, the connected subset lies on one side; relative openness would prevent any point on the other side from being a limit point of it. Therefore $L\cup V$ and $R\cup V$ are connected. They intersect in $V$, so their union $C^{\prime }$ is connected.

There is, however, no path from $(1,0)$ to $(0,1)$. If $\gamma (t)=(x(t),y(t))$ were one, let $t_0$ be the first time at which $y(t_0)=1$. Such a first time exists by continuity and compactness of $[0,1]$, and $t_0>0$. For $0\le t<t_0$, membership in the comb forces $x(t)\in \mathbb{Q}$. The continuous image of a connected space is connected, and a connected subset of $\mathbb{Q}$ is a singleton: an irrational number between any two distinct rational values would separate it. Hence $x(t)=1$ for $t<t_0$. By continuity, $\gamma (t_0)=(1,1)$, which was deleted.