I have been told that $\alpha: \mathbb{R}_\ell^2 \longrightarrow \mathbb{R} $, defined by $\alpha(x,y) = x+y$ is a quotient map. From the definition in Munkres, it needs to be the case that $U \subset \mathbb{R}$ is open if and only if $\alpha^{-1}(U) \subset \mathbb{R}_\ell^2$ is open. I am trying to understand what $U$ and $\alpha^{-1}(U)$ "look like".
From another book, I've found that a basis for $\mathbb{R}_\ell^2$ is the set of squares with side-length $\epsilon$ that is closed on the bottom and left-hand side. But I'm not sure how I can use that to obtain the set of possible $U$'s.
You don’t really have to look at all open $U\subseteq\Bbb R$: it’s enough to consider open intervals, since they form a base for the topology of $\Bbb R$. So let $U=(a,b)$ for some $a,b\in\Bbb R$ with $a<b$. Then
$$\alpha^{-1}[U]=\left\{\langle x,y\rangle\in\Bbb R_\ell^2:x+y\in U\right\}\,.$$
This is an easy set to visualize: it’s an open strip with a slope of $-1$ that intersects the $x$-axis in the interval $(a,b)\times\{0\}$ and the $y$-axis in the interval $\{0\}\times(a,b)$. Such a strip is open in the Euclidean topology on $\Bbb R^2$, which is a subset of the Sorgenfrey topology, so it’s open in $\Bbb R_\ell^2$.
Thus, if $V$ now is any open set in $\Bbb R$, $\alpha^{-1}[V]$ is a union of such strips and therefore open in $\Bbb R_\ell^2$. To complete the argument, you need only verify that every Euclidean open set in the plane is also open in $\Bbb R_\ell^2$, which follows from the fact that every Euclidean open set in $\Bbb R$ is open in $\Bbb R_\ell$.