Let $A$ be the set of real numbers that can be written in the form $\sum_{j \in J} 1/5^j$ with $J \subseteq \mathbb{N}.$ Clearly, $A$ is the set of real numbers in $[0, 1)$ whose canonical base-5 representation only contains the digits 0 and 1.
Thus, $A$ can be constructed by dividing $[0, 1)$ into 5 equal subintervals, removing the last 3 of them $[2/5, 1)$ and repeating the process over and over again with each of the remaining subintervals.
This proves that $A$ is a countable union of finite unions of intervals of the form $[a, b)$, which does not allow us to conclude that $A$ is closed or open or neither.
My intuition is that $A$ is closed. Any help proving or disproving it would be appreciated.
Let $\{0,1\}$ have the discrete topology. Since it is compact, the product $P = \prod_{j = 1}^\infty X_n$ with $X_n = \{0,1\}$ is also compact. Define $$f : P \to \mathbb R, f((x_j)) = \sum_{j=1}^\infty x_j/5^j .$$ This map is continuous. To see this, let $\epsilon > 0$. For sufficiently large $N$ we have $\sum_{j=N+1}^\infty 1/5^j < \epsilon$. Now consider $x = (x_j) \in P$. The set $U = \{x_1\} \times \ldots \times \{x_N\} \times \prod_{j = N + 1}^\infty X_n$ is an open neigborhood of $x$ in $P$. For $y \in U$ we have $$\lvert y - x \rvert = \lvert \sum_{j=N+1}^\infty (y_n -x_n)/5^j \rvert \le \sum_{j=N+1}^\infty 1/5^j < \epsilon .$$ We conclude that $f(P) = A$ is compact, hence closed in $\mathbb R$.