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$\displaystyle \bigcup_{n=2}^\infty [-1 \! + \! 1/n, \ 1) = (-1, 1)$ which is not an open set. Since the rules of topology require any union of open sets to be an open set in itself, how can $\mathbb{R}_l$ and $\mathbb{R}_k$ be considered topologies.

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    $(-1,1)$ *is* an open set in $\Bbb R_\ell$: it’s a union of basic open sets. Sets of the form $[a,b)$ are a *base* for the lower limit topology; they are not the entire topology. You might as well complain that the Euclidean topology is not a topology because $(0,1)\cup(2,3)$ is not an open interval. – 2017-01-12
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    How do you get (-1,1) otherwise. – 2017-01-12
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    The union in your question shows that $(-1,1)$ is open in $\Bbb R_\ell$: it’s a union of basic open sets, so **by definition** it’s open. – 2017-01-12

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The topology on $\mathbb{R}_l$ is usually specified by giving a base for the topology: $\mathbb{B} = \{[a,b) : a < b , a,b \in \mathbb{R}\}$, just as the usual topology is specified by giving a base as well (the open intervals, or the open balls $B(x,r)$, as for any metric space). A set $O$ is then open (by definition) iff for every $x \in O$, there is a base element $B_x$ such that $x \in B_x\subseteq O$, which amounts to saying that $O$ is a union of base elements.

For the latter definition it's easy to see that it does form a topology. The base is not the topology, but in a way the simplest open sets in the topology. If we know the base elements we can decide openness of sets.