If the ambient order is a lattice, they your order is indeed a lattice order. To see this, suppose that we have intervals (a,b) and (a',b'), in your sense that $a\le b$ and $a'\le b'$. If $b\lt b'$, then the least upper bound, written $(a,b)\vee(a',b')$, is $(b,b')$. (a'\vee b,b')$. If $b=b'$, then $(a,b)\vee(a',b')=(a\vee a',b)$. Similarly, the case $b'\le b$ is symmetric, and so the remaining case is when $b\perp b'$, in which case $(a,b)\vee(a',b')=(b\vee b',b\vee b')$. A symmetric argument shows $(a,b)\wedge(a',b')$ exists, so it is a lattice.
This argument shows that if the ambient order is an upper semi-lattice, then so is your order, and the smae for lower semi-lattice, since we only needed $\vee$ in the underlying order to define $\vee$ in your order, and the same for $\wedge$.
Also, if the underlying order is well-founded, then your order is also well-founded, since any descending sequence must descend infinitely in one of the coordinates. If the rank of the ambient order is $\omega$, then the new rank is also $\omega$, since every pair would have only have finitely many predecessors in your order. But for larger ordinal ranks, it seems to go up, and one could calculate some bounds if this was a case you were interested in.