I can't find the error in this argument, but everybody seems convinced (in the comments) that such a simply definable set with the desired property doesn't exist so there must be an error somewhere.

Let $A$ be the set of reals whose nonterminating base three expansion avoid the digit $1$. As usual, there is some hassle dealing with negative reals, so a more precise definition would be that $x \in A$ iff the nonterminating base three expansion of $2\cdot 3^k + x$ avoids the digit $1$ for all sufficiently large natural $k$. This definition makes it clear how to translate everything into the positive reals, where arithmetic is easier.

First let's check that $A$ contains no $3$-term arithmetic progression. Suppose towards a contradiction that it contains such a sequence $x < y < z$. For large $k$, look at the leftmost (most significant) digit where $2\cdot 3^k + x$ and $2\cdot 3^k + z$ differ; this is (eventually) independent of $k$. Moreover, since $x < z$, this digit must be $0$ in $2\cdot 3^k + x$ and $2$ in $2\cdot 3^k + z$. That means it must be $1$ in $2\cdot 3^k + y$ (since $y$ is the average of $x$ and $z$), contradicting the hypothesis that $y \in A$.

Now suppose that $x \not\in A$. By considering translations as above it is enough to argue in the case that $x$ is positive. Since $x$ is not in $A$, it has at least one digit which is $1$. Build distinct elements $y, z \in A$ simply by "un-averaging the $1$s," i.e., $y$ and $z$ agree with those digits of $x$ which are $0$ or $2$, and on those digits of $x$ which are $1$, make the corresponding digit of exactly one of $y,z$ equal to $0$ and the other equal to $2$. Note that if $x$ has infinitely many digits which are $1$, it's important to pick both $0$ and $2$ infinitely often for $y$ (and thus $z$) to avoid accidentally making a terminating base three expansion. Then by construction, $y, x, z$ forms an arithmetic progression in $A \cup \{x\}$.