This is probably easy, but I did not see it in standard texts.  Describe a closed subspace $V$ of $C([0,1])$ such that $V$ is a Banach lattice (in the pointwise ordering), but $V$ is not a sublattice of $C([0,1])$.  Note that, by virtue of the formula $2(f\vee g) = f+g+|f-g|$, such a $V$ must contain an element $h$ such that $|h| \notin V$.

Maybe there is a simple example using some other Banach lattice $L$ in place of $C([0,1])$, and exhibiting a closed subspace $V$ of $L$ which is a Banach lattice in the ordering inherited from $L$, but not a sublattice.