The existence of such a set follows from $``\mathbb{R}$ is a countable union of countable sets.$"$ Let $\mathbb{R} = \bigcup_{n<\omega} S_n,$ each $S_n$ countable. Let $T_n = \{x \in \mathbb{R}: \exists m \le n \exists y \in S_m (x \le_T y)\}$ and $X_n = (2^{-n-1}, 2^{-n}) \setminus T_n.$
We will show $X = \bigcup_{n<\omega} X_n$ is a subset of $(0,1)$ with the desired properties. Condition (2) follows from $X$ having cocountable intersection with each $(2^{-n}, 1).$ Suppose condition (1) fails. Let $r \in S_n$ encode a dense sequence $\langle r_i: i<\omega \rangle \subset X.$ Then $r_i \in T_n$ for all $i,$ so $2^{-n}<r_i,$ contradiction.
Edit:
It turns out the nonexistence of such a set is equivalent to $\text{AC}_{\omega}(\mathbb{R}).$ First, $\text{AC}_{\omega}(\mathbb{R})$ implies every set of reals is separable. For the other direction, suppose $\langle S_n: n<\omega \rangle$ is a sequence of nonempty sets of reals without a choice function. Let $T_n = \{x \in \mathbb{R}: \forall m \le n \exists y \in S_m (y \le_T x)\}$ and $X_n = (2^{-n-1}, 2^{-n}) \cap T_n.$
We will show $X = \bigcup_{n<\omega} X_n$ is a subset of $(0,1)$ with the desired properties. Condition (2) follows from the fact that each $T_n$ is nonempty and closed under addition by rational numbers. Suppose condition (1) fails. Let $r$ encode a dense sequence $\langle r_i: i<\omega \rangle \subset X.$ Then $\{x \in \mathbb{R}: x \le_T r\}$ is a countable set which meets each $S_n,$ which contradicts the fact that $\langle S_n \rangle$ has no choice function.