$\sqrt{2}$ is indeed a strict lower bound. A fairly short proof goes as follows:
Working with closed sets, we can assume $A$ to be a
compact set of the closed unit ball $B$ centered at $0$,
perhaps after a suitable translation. No closed ball of
radius strictly smaller than $1$ and arbitrary center
contains $A$.
By compacity of $A$, it intersects $B$ in a point
which we can assume to be the first coordinate $e_1$
of an orthonormal basis.
Suppose now that $A$ has exactly diameter $\sqrt{2}$.
Suppose first that $A$ contains no unit vector orthogonal to $e_1$. By compacity of $A$, we can
therefore assume that there exists $\epsilon>0$ such
that every element of the intersection of $A$ with the unit sphere $S$ has scalar product $\geq \epsilon$ with $e_1$. This implies that the closed halfsphere
defined by all unit-vectors making an obtuse
angle with $e_1$ is at strictly positive distance from $A$. We can therefore move the center of the unit ball
by a small amount in the direction of $e_1$ in order
to get a unit ball containing $A$ in its interior.
We can therefore assume that $A$ contains a second
element $e_2$ of an orthonormal basis. The same argument
as above shows that we can find a smaller sphere
if the $n-1$ dimensional halfsphere of unit vectors making obtuse angles with $e_1+e_2$ does not intersect $A$:
Move the center of $U$ by a tiny amount in direction $e_1+e_2$. We can therefore assume the existence
of a unit vector $e_3$ in $A$ making an obtuse angle with $e_1+e_2$. Since $A$ has diameter $\sqrt 2$, the
element $e_3$ is orthogonal to $e_1$ and $e_2$.
Iteration of
this argument shows that $A$ contains an orthonormal basis $e_1,\ldots,e_n$. But now we can apply our
favourite argument once more and move the
center of the unit ball by a tiny amount in the direction of $e_1+\ldots+e_n$ such that the moved unit
ball contains $A$ in its interior.
