Consistently, there is no such set.
Given $X\subset [0,1]^{\mathbb N}$ of cardinality continuum it suffices to find $z\in(0,1)$ such that $x\mapsto \sum_n x_n z^n$ is injective on some subset $X'\subseteq X$ of cardinality continuum. This is then more or less the same setting as [1, Theorem 2.1] where it is shown that $z$ exists in certain Cohen forcing extensions. The differences are that we want a real $z\in(0,1)$ instead of $z\in\mathbb C,$ and the functions are not entire, they're just holomorphic in the unit disc. But the proof goes through with those changes.
[1] Kumar, Ashutosh; Shelah, Saharon, On a question about families of entire functions, Fundam. Math. 239, No. 3, 279-288 (2017). ZBL1390.03044.
Also at Shelah's archive.
Under CH, there is such a set.
Assume CH. Take a Hamel basis $\{b_\beta: \beta\in\omega_1\}$ of $\ell^1(\mathbb N),$ the space of absolutely summable sequences. Define $\phi_\beta$ to be the projection by $b_\beta,$ so $\phi_\beta(x)=\sum_n b_\beta(n) x(n)$ for all $x$ in the space of bounded sequences $\ell^\infty(\mathbb N).$
Lemma. For each $\alpha\in\omega_1\setminus\omega$ there exists $x_\alpha\in[0,1]^{\mathbb N}$ such that $\phi_\beta(x_\alpha)$ is rational for $\beta<\alpha$ and irrational for $\beta=\alpha.$ (The restriction to infinite $\alpha$ is just to avoid having to treat the finite case.)
Proof. Pick a bijection $\beta:\omega\to\alpha+1.$
The functions $\phi_{\beta(n)}$ are linearly independent (as functionals on $\ell^\infty(\mathbb N)$). By linear algebra, for each $n\in\omega$ we can pick $z_n\in \ell^\infty(\mathbb N)$ such that $\phi_{\beta(m)}(z_n)=0$ for $m<n$ and $\phi_{\beta(n)}(z_n)\neq 0.$
Let $c=(\tfrac12, \tfrac12, \dots).$ Let's try a solution of the form
$$x_\alpha=c+\sum_{n=0}^\infty t_nz_n$$
with $$\max_{i}|t_nz_n(i)|\leq 2^{-n-2}\tag{1}$$ for each $n\in\omega.$ This ensures that $x_\alpha\in[0,1]^{\mathbb N}.$
Using absolute convergence to justify swapping sums, for all $n$ we have
\begin{align*}
\phi_{\beta(m)}(x_\alpha)
&=\phi_{\beta(m)}(c)+\sum_{n=0}^{\infty} t_n\phi_{\beta(m)}(z_n)\\
&=\phi_{\beta(m)}(c)+\sum_{n=0}^{m} t_n\phi_{\beta(m)}(z_n).\tag{2}
\end{align*}
In the last equality, the $t_n$ terms have vanished for $n>m$ because then $\phi_{\beta(m)}(z_n)=0.$ The coefficient $\phi_{\beta(m)}(z_m)$ of $t_m$ is non-zero. So it's easy to construct $t_m$ by induction on $m$ satisfying (1) (with $n=m$) and such that (2) is rational iff $\beta(m)\neq \alpha.$
$\square$
Pick such an $x_\alpha$ for each $\alpha\in\omega_1\setminus\omega$ and take $X=\{x_\alpha:\alpha\in\omega_1\setminus\omega\}.$ It's uncountable because the $x_\alpha$ are distinct: $\phi_\beta(x_\alpha)\neq \phi_\beta(x_\beta)$ for $\beta<\alpha.$ All the projections by basis vectors are countable: $\phi_\beta[X]\subseteq\mathbb Q\cup \{\phi_\beta(x_\alpha):\alpha\in(\beta+1)\setminus\omega\}.$ We can write an arbitrary absolutely summable sequence as $a=\sum_{\beta} \lambda_\beta b_\beta$ with reals $\lambda_\beta,$ at most finitely many non-zero. The projection by $a$ is then contained in the countable set $\{\sum_{\beta:\lambda_\beta\neq 0}\lambda_\beta q_\beta : q\in \prod_{\beta:\lambda_\beta\neq 0}\phi_\beta[X]\}.$
