Claim: Let $T_0$ be the transitive closure of $R\cup\{(a,b)\}$. Then $T_0$ is antisymmetric.
Proof.
Note that since $R$ is transitive, $x T_0 y$ means that $x R a, b R y$ or $x R y$.
Suppose $x T_0 y\land y T_0 x$.
Case 1:$x R a, b R y\land y R x$
Case 2:$x R a, b R y\land yRa,bRx$
Case 3: $xRy\land yRa,bRx$
Not possible since by transivity $bRa$.
Case 4: $xRy\land yRx$
Then $x=y$.
Is the proof valid? Should I prove $xT_0y\to(xRa,bRy)\lor xRy$?
Also a logic question: When I assume $xT_0y$ and $yT_0x$, I need to show that $x=y$ in all 4 cases. We have as a premise that $a,b$ are incomparable, so the first 3 cases are basically irrelevant, right? Since they get a contradiction against the premises, it's not a problem, but if they got a contradiction against antisymmetry, then it'd be a problem.