Distinguishability of distributions that are not close given just one sample
Total variation distance, which is exactly half of your notion of distance, is equal to
$$
d_{TV}(D,U) = \frac{1}{2} \operatorname{Distance}(D,U) = \max_A \bigl|\Pr_{x \sim D}[A(x)] - \Pr_{x \sim U}[A(x)] \bigr|.
$$
You can take $A$ to be the indicator of $\Pr[D(x)] \geq \Pr[U(x)]$. The event $A$ might be hard to compute, but it could also be easy to compute. In the latter case, you will be able to efficiently distinguish between $D$ and $U$, even using only one sample. This may even be possible in the former case, though with an advantage less than the total variation distance.

Here is a proof of the equivalence between the two formulations.
Since
$$
\Pr_{x \sim D} [A(x)] - \Pr_{x \sim U}[A(x)] =
-\left(\Pr_{x \sim D} [\overline{A(x)}] - \Pr_{x \sim U}[\overline{A(x)}]\right),
$$
where the overline represents negation, we have
$$
\delta_{TV}(D,U) = \max_A \bigl( \Pr_{x \sim D} [A(x)] - \Pr_{x \sim U}[A(x)] \bigr).
$$
Let $S = \{ x : A(x) \}$. Then
$$
\Pr_{x \sim D} [A(x)] - \Pr_{x \sim U}[A(x)] =
\sum_{z \in S} \bigl(\Pr_{Z \sim D}[Z = z] - \Pr_{Z \sim U}[Z = z]\bigr).
$$
This formula makes it clear that an optimal choice for $S$ is
$$
S = \{ z : \Pr_{Z \sim D}[Z = z] \geq \Pr_{Z \sim U}[Z = z] \}.
$$
For this choice of $S$, we have
\begin{align}
d_{TV}(D,U) &amp;=
\Pr_{x \sim D} [A(x)] - \Pr_{x \sim U}[A(x)] \\ &amp;=
\Pr_{x \sim D}[x \in S] - \Pr_{x \sim U}[x \in S] \\ &amp;=
\bigl(1-\Pr_{x \sim D}[x \notin S]\bigr) - \bigl(1-\Pr_{x \sim U}[x \notin S]\bigr) \\ &amp;=
\Pr_{x \sim U}[x \notin S] - \Pr_{x \sim D}[x \notin S].
\end{align}
By definition of $S$, we have
\begin{align}
\Pr_{x \sim D}[x \in S] - \Pr_{x \sim U}[x \in S] &amp;=
\sum_{z \in S} \Pr_{Z \sim D}[Z = z] - \sum_{z \in S} \Pr_{Z \sim U}[Z = z] \\ &amp;=
\sum_{z \in S} \bigl| \Pr_{Z \sim D}[Z=z] - \Pr_{Z \sim U}[Z=z] \bigr|.
\end{align}
Similarly, we have
$$
\Pr_{x \sim U}[x \notin S] - \Pr_{x \sim D}[x \notin S] =
\sum_{z \notin S} \bigl| \Pr_{Z \sim D}[Z=z] - \Pr_{Z \sim U}[Z=z] \bigr|.
$$
It follows that
\begin{align}
\mathrm{Distance}(D,U) &amp;= 
\sum_{z \in S} \bigl| \Pr_{Z \sim D}[Z=z] - \Pr_{Z \sim U}[Z=z] \bigr| + \sum_{z \notin S} \bigl| \Pr_{Z \sim D}[Z=z] - \Pr_{Z \sim U}[Z=z] \bigr| \\ &amp;=
\Pr_{x \sim D}[x \in S] - \Pr_{x \sim U}[x \in S] +
\Pr_{x \sim U}[x \notin S] - \Pr_{x \sim D}[x \notin S] \\ &amp;=
2\bigl(\Pr_{x \sim D} [A(x)] - \Pr_{x \sim U}[A(x)]\bigr) \\ &amp;= 2d_{TV}(D,U).
\end{align}