Given a set $S=\{1,2,3,4,5,6,7,8\}$, defined by a partial order relation Divisibility. Now consider all 4 elements containing sub-graphs, out of which $\{1,2,4,8\}$ is a Lattice obviously .
Is $\{1,2,3,6\}$ also a lattice?
Each and every pair of elements of this sub-graph has a greatest Lower Bound and Least Upper Bound Defined. So is it a Lattice?
$\{1,2,3,6\}$ is a lattice since it is the set of all divisors of $6$, and so the smallest common multiple of any of its members must also be a divisor of $6$, and likewise the greatest common divisor.
In this case the set is so small that it's easy to test this for all pairs: $\{1,2\}$, $\{1,3\}$, $\{1,6\}$, $\{2,3\}$, $\{2,6\}$, $\{3,6\}$.
If it works for pairs of members of a finite set, then a trivial induction shows that it works for larger numbers of members of that finite set. That goes like this: suppose $\{a,b\}$ has a least upper bound $c$, and $\{c,d\}$ has a least upper bound $e$. Then $e$ must be the least upper bound of $\{a,b,d\}$ since $e$ is an upper bound of $\{c\}$ and therefore of $\{a,b\}$, and is also an upper bound of $\{d\}$, and nothing smaller than $e$ can be an upper bound of $\{d,e\}$; hence cannot be an upper bound of $\{a,b,d\}$. Every time one new member is added (as $d$ was in this case) the same thing can be done again.