For which ranges of $k$ is this collection a topology on $\mathbb{R}^2$? Let $S$ be a (non-empty) subset of $\mathbb{R}$.
$$
\mathscr{T} := \left\{ \emptyset, \mathbb{R}^2 \right\} \bigcup \left\{ G_k \colon k \in S \right\},
$$
where
$$
G_k := \left\{ \, (x, y) \in \mathbb{R}^2 \, \colon \, y < x-k \, \right\}.
$$
The above collection $\mathscr{T}$ is a topology on $\mathbb{R}^2$ if $S = \mathbb{R}$ or if $S = \mathbb{N}$. Am I right?
What if $S = \mathbb{Q}$? I think then $\mathscr{T}$ won't be a topology on $\mathbb{R}^2$, because if $\left( k_n \right)_{n \in \mathbb{N} }$ be a sequence of rational numbers converging from the right to $\sqrt{2}$, for example, then we would obtain
$$
\cup_{n = 1}^\infty G_{k_n} = \left\{ \, (x, y) \in \mathbb{R}^2 \, \colon \, y < x - \sqrt{2} \, \right\},
$$
which would not be in our collection $\mathscr{T}$. Am I right?

Your reasoning is correct. Here's a more general observation.
Your sets have the following interesting property: if $x_n$ is a decreasing and convergent (in the standard, Euclidean topology) sequence then
$$G_{\lim_{n=1}^\infty x_n}=\bigcup_{n=1}^\infty G_{x_n}$$
which I leave as an exercise.

Lemma. This collection is a topology if and only if $S$ contains $\inf$ of its every subset.

Proof. "$\Rightarrow$" Assume that $A\subseteq S$, $A\neq\emptyset$ and let $s=\inf A$. If $s\in A$, then we are done. Otherwise there is a decreasing sequence $(s_n)\subseteq A$ such that $s_n\to s$. But then $G_s=G_{\lim s_n}=\bigcup G_{s_n}$ has to belong to the topology. Which means that $s\in S$.
"$\Leftarrow$" Intersection of finitely many $G_k$ is of the form $G_k$ of course. Assume that we have a collection $\{G_i\}_{i\in I}$. We want to show that the union is in the topology. Since $I\subseteq S$ then by our assumption $s=\inf I$ is in $S$ as well. I leave as an exercise that $\bigcup_{i\in I}G_i=G_s$, and thus we have a topology. $\Box$

A straight forward conclusion is that indeed your collection is a topology for $S=\mathbb{R}$ and $S=\mathbb{N}$, but not for $S=\mathbb{Q}$.
