Is this true (I can't think of counter-example). Thus, How to prove this? (assume $X$ finite with cardinality $\geq 3$, and that $B$ contains at least 2 pairs)
Intuitively, i believe the argument is just that there is some "end pair", $(a,b)$ in this case, and if $b$ was related to anything else there would be a cycle.$^*$
Writing out a formal proof though I am not clear on, particularly because there are many possible cases. So perhaps induction is the way to go? (but then induction on what?)
Some more of my thoughts: Maybe try induction on the number of pairs in $B$?
Suppose $B$ has two pairs. Then they must be of the form $(a,b)$ and $(c,d)$ . If $c=b$ then $d\not=a$ so $(d,y)\not \in B$ for all $y$ and the claim is satisfied. the $d=a$ case is similar, and if $c\not=b, d\not =a$, then both pairs satisfy the claim.
Suppose it is true when $B$ has $n$ pairs and consider when $B$ has $n+1$ pairs. Somehow reduce this to a relation on $n$ pairs?
I'm stuck, and I don't know if this is the right approach even.
$^*$that is, $(a,b)$ implies NOT $(b,a)$, and then if $(b,c)\in B$, the NOT $(c,a)$ or $(c,b)$. i.e., the longer the "chain" the more restrictions, so we should either end up with a "chain" that uses all elements, or the chain ends prematurely which means the premise is satisfied by whatever the last element in the "chain" is. (I use "chain" informally to mean something like $(a,b), (b,c), (c,d) \in B$ because I can't recall a technically term for that.
By contradiction, let's $B$ is a such acyclic binary relation on $X$ that for all $(a,b)\in B$ there is $c\in X$ such that $(b,c) \in B$.
Consider some pair $(x_1,x_2)\in B$ (there is at least one since $B$ is nonempty). There is some $x_3 \in X$ such that $(x_2,x_3) \in B$. Therefore there is some $x_4 \in X$ such that $(x_3,x_4)\in B$, etc. So we have a "chain" of pairs in $B$ $$ (x_1,x_2),\, (x_2,x_3),\, (x_3,x_4),\, \dots $$ Now if $|X| = n$ we have $x_{n+1} = x_k$ for some $k \le n$ (otherwise there is at least $n+1$ elements in $X$), but then we have a cycle $$ (x_k, x_{k+1}),\,\dots, (x_{n-1},x_n),\, (x_n, x_k) $$ in $B$, which we supposed to be acyclic.