The topology a line inherits as a subspace of $\mathbb{R}_l \times \mathbb{R}$, or of $\mathbb{R}_l \times \mathbb{R}_l$ (Munkres) Exercise 8, Chapter 2, Section 16, Page 92, from the Munkres's book:
The topology $\mathbb{R}_l$ is the topology on $\mathbb{R}$ generated by all half-open intervals of the form
\begin{align*}
[a, b) = \{ x \, |\, a \le x < b \},
\end{align*}
where $a < b$.
By Lemma 13.4 from the Munkres's textbook, this topology is finer than the standard topology on $\mathbb{R}$. Thus, all open sets of the standard topology on $\mathbb{R}$ also belong to the topology of $\mathbb{R}_l$. By Theorem 15.1 from the Munkres's textbook, a basis for the product topology $\mathbb{R}_l \times \mathbb{R}$ is the collection of all the products $[a, b) \times (c, d)$, where $[a, b)$ is a half-open interval in $\mathbb{R}$, and $(c, d)$ is an open interval in $\mathbb{R}$. The products $[a, b) \times (c, d)$ represent the interior of rectangles, where the rectangles have sides parallel to the coordinate axes, along with the left side of the rectangle. The corners do not belong to the open set.
To avoid difficulty with notation, we shall denote the general element of $\mathbb{R} \times \mathbb{R}$ by $x \times y$. Let $L$ be a straight line in the plane $\mathbb{R}^2$; then, $L$ is a subset of points of $\mathbb{R} \times \mathbb{R}$. The topology $L$ inherits as a subspace of $\mathbb{R}_l \times \mathbb{R}$, has as basis all sets of the form $[a, b) \times (c, d) \cap L$; it is easy to see that the intersection is a line segment of $L$. We have three possible cases:
(1) The straight line $L$ is parallel to $x$-axis. Therefore, $L$ intersects the left side of the rectangles, and then the points of the line segment belong to a half-open interval of $L$, which can be denoted as $[\alpha, \beta)$, where $\alpha,\beta \in L$. We deduce that the topology $L$ inherits as subspace of $\mathbb{R}_l \times \mathbb{R}$ is the topology of $\mathbb{R}_l$.
(2) The straight line $L$ is parallel to $y$-axis. Therefore, $L$ intersects the upper and lower sides of the rectangles, and then the points of the line segment belong to an open interval of $L$, which can be denoted as $(\alpha, \beta)$, where $\alpha,\beta \in L$. Therefore, the topology $L$ inherits as subspace of $\mathbb{R}_l \times \mathbb{R}$ is the standard topology on $\mathbb{R}$.
(3) The straight line $L$ is not parallel to any of the axes. Therefore, if $L$ intersects the left side of the rectangles, the points of the line segment belong to a half-open interval of $L$, which can be denoted as $[\alpha, \beta)$, where $\alpha,\beta \in L$. But if $L$ does not intersect the left side of the rectangles, the points of the line segment belong to an open interval of $L$, which can be denoted as $(\alpha, \beta)$, where $\alpha,\beta \in L$. The collection of half-open intervals $[\alpha, \beta)$ is a basis for the topology of $\mathbb{R}_l$, and the open intervals $(\alpha, \beta)$ are open sets in $\mathbb{R}_l$. It follows that the topology $L$ inherits as subspace of $\mathbb{R}_l \times \mathbb{R}$ is the topology of $\mathbb{R}_l$.
Now, let us consider the topology $L$ inherits as a subspace of $\mathbb{R}_l \times \mathbb{R}_l$. By Theorem 15.1 from the Munkres's textbook, a basis for the product topology $\mathbb{R}_l \times \mathbb{R}_l$ is the collection of all the products $[a, b) \times [c, d)$, where $[a, b)$ is a half-open interval in $\mathbb{R}$, and $[c, d)$ is a half-open interval in $\mathbb{R}$. The products $[a, b) \times [c, d)$ represent the interior of rectangles, where the rectangles have sides parallel to the coordinate axes, along with their left and lower sides. Note that the lower left corner of the rectangle belongs to the product $[a, b) \times [c, d)$ as well.
Let $L$ be a straight line in the plane $\mathbb{R}^2$. The topology $L$ inherits as a subspace of $\mathbb{R}_l \times \mathbb{R}_l$, has as basis all sets of the form $[a, b) \times [c, d) \cap L$. We have also three possible cases:
(1) The straight line $L$ is parallel to $x$-axis, or is parallel to $y$-axis. In the first case, $L$ intersects the left side of the rectangles, and in the second case, $L$ intersects the lower side of the rectangles. In both cases, the points of the line segment belong to a half-open interval of $L$, which can be denoted as $[\alpha, \beta)$, where $\alpha,\beta \in L$. We deduce that the topology $L$ inherits as subspace of $\mathbb{R}_l \times \mathbb{R}_l$ is the topology of $\mathbb{R}_l$.
(2) The straight line $L$ is not parallel to any of the axes, and makes an angle less than $\pi / 2$ with the $x$-axis. Therefore, $L$ intersects the left side, the lower side, or the lower left corner of the rectangles, and then, the points of the line segment belong to a half-open interval of $L$, which can be denoted as $[\alpha, \beta)$, where $\alpha,\beta \in L$. Again, the topology $L$ inherits as subspace of $\mathbb{R}_l \times \mathbb{R}_l$ is the topology of $\mathbb{R}_l$.
(3) The straight line $L$ is not parallel to any of the axes, and makes an angle greater than $\pi / 2$ with the $x$-axis. Therefore, $L$ may intersect any sides of the rectangles, or may only intersect the lower left corner of the rectangle. Then, if the intersection is a line segment of the straight line $L$, the points belong to an interval open, half-open or closed of $L$; if the intersection is a point of $L$, we have a one-point open set (singleton set). Then, the topology $L$ inherits as subspace of $\mathbb{R}_l \times \mathbb{R}_l$ is the discret topology on $\mathbb{R}$.
 A: This problem is actually rather long and complicated, since the answer in each case depends on the slope of $L$. However, it is true that in each case it is a familiar topology, meaning one that you’ve encountered before: if $\mathscr T$ is the induced topology on $L$, $\langle L,\mathscr T\rangle$ will in every case be homeomorphic to $\langle\Bbb R,\mathscr T_0\rangle$ for some topology $\mathscr T_0$ on $\Bbb R$ that you’ve seen before.
I’ll work one of the cases and sketch another to give you an example of what you have to do. Suppose that $L$ is a line of negative slope, and we consider it as a subspace of $\Bbb R_\ell\times\Bbb R_\ell$. If $p=\langle x,y\rangle$ be any point of $L$, let 
$$U=[x,x+1)\times[y,y+1)\;;$$
clearly $U$ is open in $\Bbb R_\ell\times\Bbb R_\ell$, and $U\cap L=\{p\}$, so $\{p\}$ is open in $L$, and the induced topology on $L$ is the discrete topology. This is familiar even without relating it to $\Bbb R$, but in fact you can easily show that the map $$\varphi:L\to\Bbb R:\langle x,y\rangle\mapsto x$$ is a homeomorphism from $L$ with the subspace topology to $\Bbb R$ with the discrete topology.
Now suppose that $L$ has positive slope, and again consider it as a subspace of $\Bbb R_\ell\times\Bbb R_\ell$. Let $\pi:L\to\Bbb R_\ell:\langle x,y\rangle\mapsto x$. It’s easy to show that $\pi$ is a continuous bijection, so to prove that it’s a homeomorphism you need only show that it’s an open map. We know that the sets of the form $[a,b)\times[c,d)$ with $a<b$ and $c<d$ form a base for the topology of $\Bbb R_\ell\times\Bbb R_\ell$, so the intersections of these sets with $L$ form a base for the subspace topology on $L$. Let $U=[a,b)\times[c,d)$, and suppose that $U\cap L\ne\varnothing$. Then either there is an $x_0\in[a,b)$ such that $$U\cap L=\{\langle x,y\rangle\in L:x_0\le x<b\}\;,$$ or there is a $y_0\in[c,d)$ such that $$U\cap L=\{\langle x,y\rangle\in L:y_0\le y<d\}\;.$$ In the first case it’s easy to see that $\pi[U\cap L]$ is open in $\Bbb R_\ell$. Probably the most straightforward way to show that the same is true in the second case is to let $y=mx+r$ be the equation of $L$, where $m>0$, and see what the inequality $y_0\le y<d$ tells you about $x$.
You still have to consider the cases in which $L$ is horizontal or vertical, and all of the cases when the ambient space it $\Bbb R_\ell\times\Bbb R$.
A: He presumably means that you've encountered the resulting topologies before.
For any line $L$ in the plane, one way to define a topology on it is to project the line onto the $x$- or $y$-axis (always at least one of these will be bijective, and so always make such a choice), so we can identify the line with $\mathbb{R}$. Put another way, if we call the projection $\pi$ and have a topology $\tau$ on $\mathbb{R}$ we can always define a topology on $L$ by declaring a set $U \subset L$ to be open iff $\pi(U)$ is open in $\tau$, so surely Munkres means that the topology on $L$ arises that from a familiar topology on $\mathbb{R}$ in this way.
