I tried to prove the following: There exists a set $A\subseteq \mathbb{R}$ such that $|A|>\aleph_0$, and, for every $x_1,\ldots,x_n\in A$, $$x_1+\ldots+x_n \notin \mathbb{Q}$$
I found a solution using Zorn's lemma. I'm interested to know whether there exists a solution which doesn't apply the axiom of choice.
My solution: Apply Zorn's lemma to find a maximal element among subsets $A\subseteq \mathbb{R}$ such that any finite sum of elements from $A$ doesn't belong to $\mathbb{Q}$. Now, if the maximal $A$ was countable, then we could extend it to $A\cup \{r\}$, where $r\in \mathbb{R}$ is picked as follows: take $r\in \mathbb{R}$ which doesn't belong to the countable set of elements of the form $q-S$, where $q\in \mathbb{Q}$ and $S$ is a sum of finitely many elements of $A$.
Yes: You can take $$ A=\biggl\{ \sum_{n=0}^\infty 10^{-n^2} f(n) \biggm| f \in \mathbb \{1,2\}^{\mathbb N} \biggr\}$$ This clearly has $2^{\aleph_0}$ elements. And the nonzero digit in the decimal expansions are ever further from each other, so for every finite sum of elements from the set, there will be increasingly long runs of $0$s in the decimal expansion of the sum ...
If we change the exponents in the scaling factor $10^{-n^2}$ to grow fast enough (something like $10^{-n!}$ should suffice), we can even make sure that all of the finite sums will be Liouville numbers and therefore transcendental.