The smallest distance is 3, achieved by $A = (0,0)$, $B = (3,0)$, $C = (0,4)$.
To check this, it suffices to check that $AB$ cannot equal 1 or 2. (It cannot equal 0
because if two of the points were to coincide, the three points would be collinear.)

The triangle inequality implies that $|AC - BC| \leq AB$, with equality if and only if $A,B,C$
are collinear. If $AB = 1$, we may assume without loss of generality that $A = (0,0)$, $B = (1,0)$.
To avoid collinearity, we must have $AC = BC$, but this forces $C = (1/2, y)$ for some $y \in \RR$,
a contradiction. (One can also treat this case by scaling by a factor of 2 to reduce to the case $AB=2$,
treated in the next paragraph.)

If $AB = 2$, then we may assume without loss of generality that $A = (0,0), B = (2,0)$.
The triangle inequality implies $|AC - BC| \in \{0,1\}$.
Also, for $C = (x,y)$, $AC^2 = x^2 + y^2$ and $BC^2 = (2-x)^2 + y^2$ have the same parity;
it follows that $AC = BC$. Hence $c = (1,y)$ for some $y \in \RR$, so $y^2$ and $y^2+1=BC^2$
are consecutive perfect squares. This can only happen for $y = 0$, but then $A,B,C$ are collinear,
a contradiction again.