Suppose that in the Cartesian plane $\mathbb{R}^2$ we let $X$ denote the union of all lines through the origin with rational slope. Would that make $X$ connected, since all such lines are connected and only intersect at the origin, i.e. $(0,0)$? Would this then mean that $X- \{(0,0)\}$ is not connected? Would it even have any connected components, and if so, what could they be and why?
For my second question, again in the Cartesian plane, suppose we let $X$ be the union of horizontal lines defined in the following manner: First, this space includes the closed unit interval, or perhaps even any such interval like $[p,q]$. Next, we have a line $L_1$ of the same length as $[p,q]$, i.e. $q-p$, whose vertical distance (height) from $[p,q]$ is $1$. Then, we have a second line $L_2$ again of length $q-p$, but this time with height $1/2$ from $[p,q]$. Continuing this countable collection, we have line $L_i$ of length $q-p$ with height $1/i$ from $[p,q]$. Note that all these lines are parallel to $[p,q]$. What would be the connected components in this case, and how come?
For the first question, since connected components are closed and connected, my guess would be that the only connected component would be $\mathbb{R}^2 - \{(0,0)\}$, but I might be wrong. For the second, again I'm not sure of this, but I think since all of the $L_i$ are disjoint, the only connected component would be that closed interval that lies on the $x$-axis. I would really appreciate some help here.