Comb space: cannot understand why it is not locally path-connected

Question

Wikipedia says that the comb space is path-connected but not locally path-connected. I cannot see this. Even when I read here that at any point in $\{0\}\times (0,1]$ it is not locally path-connected, I still cannot see it. For example, if we took the point $p=(0,1)$, surely the neighbourhood $\{0\}\times (1-\epsilon,1]$ is open and path-connected, and contains $p$, and if we make $\epsilon$ small enough we can get it to fit inside any open neighbourhood of $p$.

Answer 1

The error is always near words like "surely" ;)

The set $A:=\{0\}\times (1-\epsilon,1]$ is not open in the comb space $C$. Indeed $((1/n, 1))_{n\in\mathbb N}$ is a sequence in $C\setminus A$ which converges to $(0,1)\in A$.