Another quick solution: use two letters ($\Sigma = \{a,b\}$) and use $b$ only as a separator, then make $L$ irregular using a counting argument in such a way that in $L \cup L^R$ the counting argument "disappears" ...

Solution:

>! if $L = \{a^nba^m | n \geq m \}\;$, then  $L^R =\{a^nba^m | n < m \} $<br> and therefore $L \cup L^R = \{a^*ba^*\}$