Let us consider the Thales circle over $BY$, which is circumscribed to $BYX$. This circle contains $X$ and touches $AB$ in $B$. Of all such circles, the one which touches $AE$ (and it has to be in $X$) obviously has the least diameter (let us call the circle $k$). This circle is thus inscribed to the equilateral triangle $AA'F$ where $A'$ is the image of $A$ in the point symmetry with respect to $B$ and $F$ lies on the half line $BC$ (see Fig. 1; there you can see one of the circles with smaller diameter than $k$ as well). The center of $k$ is the center of mass of the triangle $AA'F$, equilateral triangle with sides of length $2$, thus the diameter of $k$ is $BY = \frac{2}{3}\sqrt{3}$ and the corresponding $X$ is a center of $AF$, that is it belongs to the segment $AE$, since $|AX| = 1 < |AE|$.

Answer. The least possible length of $BY$ is $\frac{2}{3}\sqrt{3}$.