Although your setup is in the interval $[0,1)$, I will ignore the left endpoint and work with $(0,1)$. Recall the cumulative distribution function for the normal distribution:
$$
\Phi(t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^t e^{-(1/2)x^2}dx.
$$
This function is an increasing map from ${\mathbb R}$ onto $(0,1)$ and its value at $t$ gives the probability that a normal random variable has value less than or equal to $t$.
Let $g \colon (0,1) \rightarrow {\mathbb R}$ be its inverse, i.e., $g(y)$ is the unique solution $t$ to $\Phi(t) = y$. That is,
$$
\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{g(y)} e^{-(1/2)x^2}dx = y.
$$
Note the input to $g$ is a number in $(0,1)$ and the output is a real number. That's the kind of setup you're looking for.

Claim: If $X$ is a uniform random variable on $(0,1)$ then $g(X)$ is a normally distributed random variable on the real line.

Proof: For $a < b$ in ${\mathbb R}$, we want to show the probability $a \leq g(X) \leq b$ is
$$
\frac{1}{\sqrt{2\pi}}\int_{a}^b e^{-(1/2)x^2}dx = \Phi(b) - \Phi(a).
$$
Since $g$ and $\Phi$ are inverses of each other, the condition $a \leq g(X) \leq b$ is the same as $\Phi(a) \leq X \leq \Phi(b)$, which is a condition in $(0,1)$ and your hypothesis about $X$ being *uniform* in $(0,1)$ is that the probability of that is $\Phi(b)- \Phi(a)$. Since
$$
\Phi(b) - \Phi(a) = \frac{1}{\sqrt{2\pi}}\int_{a}^b e^{-(1/2)x^2}dx,
$$
we've got a Gaussian distribution. (That you want $X$ to be uniform made this a lot easier to describe.)

Quite generally, if you want to model a probability distribution on the real line with density function $f(x)$ by sampling a uniform random variable $X$ on $(0,1)$, you can use
the function $g(X)$, where $g$ is the inverse of the cumulative distribution function
$F(t) = \int_{-\infty}^t f(x)dx$.