Is every open subset of $ \mathbb{R} $ uncountable?

The empty set.

Otherwise, yes. Every open interval is uncountable, so every nonempty open subset of $\Bbb R$ is uncountable.

I assume you don't want a proof but hints. To prove all intervals are uncountable you could first try proving (0,1) is uncountable. Can you construct a decimal fraction of a number that is in (0,1) but isn't in a supposed sequence of all numbers in (0,1)?

After you've established (0,1) is uncountable you could try constructing a bijection between (a,b) and (0,1) and thus proving (a,b) is uncountable.

Every open subset of $\mathbb{R}$ is a nonempty union of disjoint open intervals, each of which are open. A union of open sets is always open. Every nonempty open interval is uncountable.

For any point $x$ of an open set $S$, $S$ contains a neigbourhood of $x$, and this neighbourhood has uncountably many elements. So as long as $x\in S$ exists (in other words $S\neq\emptyset$) $S$ is uncountable.