If $A\subseteq \mathbb{R}$ and $A$ has no isolated points, then $A$ is uncountable

Question

I want to show that a set $A\subseteq \mathbb{R}$ is uncountable if it has no isolated points. I have seen on other posts that this is true if $A$ is a complete metric space with no isolated points, but I don't know if this result applies here? $A$ can of course be considered a metric space with the restriction of the metric to $A$, but I don't think it holds in general, that subsets of complete spaces are complete.
I don't know too much about topology, so I'm not sure where to begin with this if the result I just mentioned does not apply here.

Answer 1

$\Bbb{Q}$ is a subset of $\Bbb{R}$ which has no isolated points (it is actually a dense subset) but it is countable. And no, subsets of complete metric spaces are not necessarily complete ($\Bbb{Q}$ being an example, or even simpler, $(0,1)$). However, closed subsets, equipped with the induced metric, of complete metric spaces are again complete (but $\Bbb{Q}$ is not closed in $\Bbb{R}$, so again this theorem can't be applied).