Open set in irrationals

Let $\Bbb P=\Bbb R\setminus\Bbb Q$ be the set of irrationals. Let $U$ be a non-empty open set in $\Bbb R$; then there are $a,b\in\Bbb R$ such that $a<b$ and $(a,b)\subseteq U$. As you say, the rationals are dense in $\Bbb R$, so there is a rational $q\in(a,b)$, and it follows that $$q\in(a,b)\setminus\Bbb P\subseteq U\setminus\Bbb P$$ and hence that $U\nsubseteq\Bbb P$. Thus, as you suspected, the irrationals do not contain any non-empty open subset of $\Bbb R$.

The fact that $\Bbb Q$ is countable and therefore can be covered by an open set of small radius is simply irrelevant: there is no reason to think that in general an open set covering a dense set should cover the whole space. A dense set can even be open. For a very simple example, consider the subspace

$$X=\{0\}\cup\left\{\frac1n:n\in\Bbb Z^+\right\}$$

of $\Bbb R$: the set $X\setminus\{0\}$ is a dense, open subset of $X$, so it is an open cover of itself that covers only itself, not all of $X$.