Let $\Gamma$ be the closed loop the ant crawled. Let $A,B\in \Gamma$ be two points such that the length of the path that the ant travels from $A$ to $B$ is exactly $1/2.$ Then the rest of the curve (from $B$ to $A$) has also length $1/2$. Thus, we have two paths $P_1, P_2$ from $A$ to $B$, each of length $1/2$. Clearly $|AB|\le 1/2.$ Take two points $A_1, B_1$ lying on the line $AB$ such that $B$ is between $A$ and $B_1,$ and $A$ is between $A_1$ and $B.$ Let also $|AB_1|+|B_1B|=|BA_1|+|A_1A|=1/2.$ Let $\Gamma':= \{X : |AX|+|XB|=1/2\}.$ It means $\Gamma'$ is an ellipse with focuses $A$ and $B.$ Note that $P_1$ and $P_2$ lie inside $\Gamma'.$ Indeed, assume there is a points $Y\in P_i$ that is outside $\Gamma'.$ Then $|AY|+|YB|>1/2$ which means the length of $P_i$ is more than $1/2$, contradiction. It remains to see that $\Gamma'$ lies inside the circle with center the midpoint of $AB$ and radius $1/2.$