Is $\mathbb Q \subset \mathbb R$ the subspace topology, discrete?, and questions about showing discreteness

You can show a (metric) space is not discrete by producing a sequence that converges to a point that doesn't appear in the sequence.

This relies on the fact that a sequence in a discrete space is convergent if and only if it is eventually constant.

The example for $\mathbb{Q}$ is the sequence $1/n$.

$\mathbb{Q}$ is definitely not discrete in $\mathbb{R}$. Suppose that we have some open set $O$ of the reals such that $O \cap \mathbb{Q} = \{q\}$ for some $q \in \mathbb{Q}$ (which should exist if $\{q\}$ were open). Then as $q \in O$ we have some open interval $(q-t, q+t) \subseteq O$. Between any two distinct reals lies a rational number (and an irrational number too), so we have some $q' \in \mathbb{Q}$ with $q < q' < q+t$. But then this $q' \in O \cap \mathbb{Q}$ too, so this set cannot be equal to just $\{q\}$, contradiction. No singleton is thus open in the subspace topology on $\mathbb{Q}$, every point of $\mathbb{Q}$ is a limit point of $\mathbb{Q}$. The same holds for the irrationals $\mathbb{P}$ in the subspace topology.

Besides the classical $\mathbb{N}$ and $\mathbb{Z}$ which are discrete we have $A = \{\frac{1}{n}: n=1,2,\ldots \}$ as a discrete subspace (although $0$ is a limit point of $A$ so $A$ is not closed, no point of $A$ itself is a limit point of $A$). In any metric space all finite subspaces are discrete.