This is a standard application of Baire's Theorem (BCT3):
Assume that indeed
$$
(0,1)\cap\mathbb Q=\bigcap_{n\in\mathbb N}U_n,
$$
where $U_n$ open dense in $[0,1]$, and $(0,1)\cap\mathbb Q=\{q_n\}_{n\in\mathbb N}$.
Now define
$$
V_n=U_n\smallsetminus\{q_n\}.
$$
Then clearly, $V_n$ open and dense in $[0,1]$, and
$$
\bigcap_{n\in\mathbb N}V_n=\varnothing,
$$
which contradicts Baire's Theorem, as $[0,1]$ is a complete metric spaces, and an intersection of countably many open and dense subsets of it is empty.
EDIT. If you do not want to use Baire's Theorem, let $U_n$'s and $V_n$'s defined as above. As $V_1$ is open, then there is a non-empty interval $(a_1,b_1)$, such that
$$
(a_1,b_1)\subset V_1\subset (0,1).
$$
Choose now a closed interval $[c_1,d_1]\subset (a_1,b_1)$, with $d_1>c_1$.
Next, as $V_2$ is open and dense, then there is a non-empty interval $(a_2,b_2)$, such that
$$
(a_2,b_2)\subset V_2\cap (c_1,d_1),
$$
and choose a closed interval $[c_2,d_2]\subset (a_2,b_2)$, with $d_2>c_2$.
Recursively we can thus obtain two sequences of intervals $(a_n,b_n)$, $[c_n,d_n]$, $n\in\mathbb N$, such that
$$
[c_{n+1},b_{n+1}]\subset (a_{n+1},b_{n+1})\subset V_{n+1}\cap (c_n,d_n).
$$
But $\bigcap_{n\in\mathbb N} V_n=\varnothing$ implies that $\bigcap_{n\in\mathbb N} [c_n,d_n]=\varnothing$, which is a contradiction.
Best Answer
$[0, 1]$ is a complete metric space so we can apply the Baire category theorem to it. If we had such a countable intersection $[0, 1] \cap \mathbb Q = \bigcap U_i$ then each $V_i = U_i \cap [0, 1]$ would be an open subset of $[0, 1]$ containing $\mathbb Q$. Hence, each $V_i$ is open and dense in $[0, 1]$. Furthermore, we'd still have $[0, 1] \cap \mathbb Q = \bigcap V_i$
On the other hand, let $\{q_i\}_{i \in \mathbb N}$ be an enumeration of $\mathbb Q \cap [0, 1]$. Then $V_i - \{q_i\}$ will still be open and dense in $[0, 1] \cap \mathbb Q$, so the intersection $\bigcap (V_i - \{q_i\})$ must be dense in $[0, 1]$ by the Baire category theorem. However, $\bigcap (V_i - \{q_i\}) = (\bigcap V_i) - [0, 1] \cap \mathbb Q = \emptyset$. But the empty set isn't dense in $[0, 1]$ - contradiction. Hence, $\mathbb Q \cap [0, 1]$ is not a $G_\delta$.