Why is the deleted comb space still connected? I read from wikipedia that the deleted comb space:
$$
(\{0\} \times \{0,1\}) \bigcup (K \times [0,1]) \bigcup ([0,1] \times \{ 0 \})
$$
has a well known result of being connected.. But how is it connected if there's a separate point $p=<0,1>$? Or is it because since $p$ is a point and is not considered to be among the open sets of $\mathbb{R}$ which are in the form of intervals...

Clearly, if $Y=\bigl(K\times[0,1]\bigr)\cup\bigl([0,1]\times\{0\}\bigr)$, then $Y$ is path connected, and therefore connected. But your set lies between $Y$ is its closure. Therefore, it is connected too.
