Solution:
Answer: $2/3$.

Let $O$ be the center of $ABC$. Let $AO$ meet $BC$ at $D$, let $BO$ meet $CA$ at $E$, and let $CO$ meet $AB$ at $F$. Given any point $X$ inside $ABC$, it lies in one of the quadrilaterals $AEOF$, $CDOE$, $BFOD$. Without loss of generality, it lies in $AEOF$. Take the line through $X$ parallel to $BC$. It meets $AB$ in $P$ and $AC$ in $Q$. Then $PQ$ is shorter than the parallel line $MON$ with $M$ on $AB$ and $N$ on $AC$, which has length $2/3$.

If we twist the segment $PXQ$ so that it continues to pass through $X$, and $P$ remains on $AB$ and $Q$ on $AC$, then its length will change continuously. Eventually, one end will reach a vertex, whilst the other will be on the opposite side and hence the length of the segment will be at least that of an altitude, which is greater than $2/3$. So at some intermediate position its length will be $2/3$.

To show that no value smaller than $2/3$ is possible, it is sufficient to show that any segment $POQ$ with $P$ and $Q$ on the sides of the triangle has length at least $2/3$. Take $P$ on $MB$ and $Q$ on $AN$ with $P$, $O$, $Q$ collinear. Then $PQ \cos POM = MN - QN \cos \pi/3 + PM \cos \pi/3$. But $PM > QN$ (using the sine rule, $PM = OM \sin POM/\sin OPM$ and $QN = ON \sin QON/\sin OQN$, but $OM = ON$, $\angle POM = \angle QON$, and $\angle OQN = \angle OPM + \pi/3 > \angle OPM$), and hence $PQ > MN \sec POM > MN$.