Let $S \subset \mathbb R$, and fix an $x \in S\backslash\mathbb Q$. Then is the following statement valid?
There exists an $\epsilon > 0$ such that $\vert{x - y}\vert > \epsilon$ for all $y \in \mathbb Q \cap S$.
An obvious candidate for $\epsilon$ would be $\inf \{\vert x-y \vert : y \in \mathbb Q \cap S \}$. But can we show that this value is strictly positive?
On
Note that while the other answer is technically correct, it doesn't address your question because there are no irrationals in $\{\pi_n\}$. In other words, there's no $x$ to choose.
It's a property of the rationals that you can approximate irrationals arbitrarily well. Take $S=\{e(1+(-1)^n)-2-\sum\limits_{1\le k\le n} 2/k! \big\vert n\in \mathbb{N}\}.$ We can only pick $x=2e-2$ (we get it when $n=0$, by definition of empty sum); on the other hand $\sum_{k\ge1} 2/k!=2e-2=x$, meaning that you can't find an $\varepsilon$ that works for all rational $y$ in $S$.
No, you can't. For a given irrational number (say $\pi$), it has a decimal expansion. We have that the sequence of rational numbers $\lbrace x_n\rbrace = (3,3.1,3.14,\dots)$ has the property that, for any $\epsilon>0$, that for all $n\geq N$ for some $N$ that $|\pi-x_n|<\epsilon$. This argument can be repeated for any irrational number's decimal expansion.
The proof that the $N$ exists isn't that hard. We have that for any $\epsilon>0$ that for some $k$ there is $10^{-k}<\epsilon$ (by the archimedean property). So, for all $n\geq k$, we'll have the result.