What Topology does a Straight Line in the Plane Inherit as a Subspace of $\mathbb{R_l} \times \mathbb{R}$ and of $\mathbb{R_l} \times \mathbb{R_l}$

You’re right about $\Bbb R_l\times\Bbb R$, but you can make a better (or at least clearer) argument. Basic open sets in $\Bbb R_l\times\Bbb R$ are of the form $[a,b)\times(c,d)$, so they’re boxes open on all sides except the left. Let $L$ be a non-vertical line, and let $\langle x,y\rangle\in L$. Then the intersection of any basic open set $[x,b)\times(c,d)$, where $y\in(c,d)$, is a left-closed, right-open interval with $x$ as left endpoint. It’s not hard to check that these intersections are a base for the subspace topology on $L$, which is therefore the Sorgenfrey (lower-limit) topology. And as you say, the vertical lines are homeomorphic to the second factor, $\Bbb R$.

$\Bbb R_l^2$ is another story, though. Now the topology has a base of boxes of the form $[a,b)\times[c,d)$, open only on the top and righthand edges. If the line $L$ is vertical or horizontal, of course, it’s homeomorphic to one of the factor spaces, i.e., to $\Bbb R_l$. If it has positive slope, you can argue much as in the case of $\Bbb R_l\times\Bbb R$ to conclude that it’s homeomorphic to $\Bbb R_l$. If it has negative slope, however, its topology is discrete: for any $\langle x,y\rangle\in L$, $$L\cap\Big([x,x+1)\times[y,y+1)\Big)=\big\{\langle x,y\rangle\big\}\;.$$

I wish to respond to @lomber's question to Brian M. Scott's answer, however I don't have enough Math Exchange points to do so directly. With that, I answer it here:

Notice that, if $\langle x,y\rangle\in L $, then

$$\{\langle x,y\rangle\}\stackrel{(1)}{=}L\cap([x,x+1)\times[y,y+1))$$ is an open set in the subspace topology on $ L $, inherited by $ \mathbb{R}_\ell^2 $. This tells us that for every element $ \langle x,y\rangle\in L $, the set $ \{\langle x,y\rangle\} $ is open. Thus, if $ U $ is any subset of $ L $, then $$U=\bigcup_{\langle x,y\rangle\in U}\{\langle x,y\rangle\}$$ is an open set, being a union of open sets. This informs us every subset of $ L $ is open. This is the definition of the discrete topology.

To justify equation (1) (if that was the difficult part), notice if $ L $ has a negative slope and $ \langle x,y\rangle\in L $ (i.e. $y=mx+b$ with $ m<0 $), then $ v>x$ implies $$mv+b<mx+b=y\text{ thus }mv+b\notin[y,y+1) $$ so that $\langle v,mv+b\rangle\in L\backslash\big([x,x+1)\times[y,y+1)\big). $ Otherwise, $ v<x $ clearly gives $ \langle v,mv+b\rangle\in L\backslash\big([x,x+1)\times[y,y+1)\big)$.

Thus if $$\langle v,w\rangle\in L\cap([x,x+1)\times[y,y+1)) $$ and $\langle x,y\rangle\in L $, then $ v=x $ so that $ w=y $, whence $$L\cap([x,x+1)\times[y,y+1))\subset\{\langle x,y\rangle\}.$$ The reverse inclusion comes from $ \langle x,y\rangle\in L $.