Solution:

Picture an instant in time where the ant and spider have $x$-coordinates $a$ and $s$, respectively. If $1 \leq s \leq 3$, then $a \leq 0$, and the distance between the bugs is at least $1$. If $s > 3$, then, needless to say, the distance between the bugs is at least $2$. If $-1 \leq s \leq 1$, then $s = 1 - 2a$, and the distance between the bugs is
$$\sqrt{(a - (1 - 2a))^2 + (1 - a^2)} = \sqrt{8a^2 - 6a + 2} = \sqrt{\frac{(8a - 3)^2 + 7}{8}}$$
which attains the minimum value of $\sqrt{7/8}$ when $a = 3/8$.