To construct a lattice out of a chain (K,<=), first define
. . . x min y = x if x <= y, = y if y <= x,
. . . x max y = y if x <= y, = x if y <= x.
Since K is a chain, these definitions are well defined for all x,y.
To show x min y is the greatest lower bound of A = { x,y },
assume b is a lower bound of A. Thus b <= x,y and either
x <= x min y or y <= x min y. Either way b <= x min y.
Thus x min y is the glb of A.
In a similar manner, x max y is the least upper bound of A.
Consequently (K,min,max) is a lattice.
Since a = a min b iff a <= b iff b = a max b
one can, if desired, prove all the equations
for an algebraic description of a lattice.
To show (K,min,max) is a distributive lattice both
. . a min (x max y) = (a min x) max (a min y)
and
. . a max (x min y) = (a max x) min (a max y)
are to be proven.
Before proceeding with the proof, show that if x <= y,
then a min x <= a min y and a max x <= a max y.
For the first equation, there are two cases: x <= y and y <= x
which by symmetry, only one needs to be considered, that is wlog.
So assume x <= y. Whence
(a min x) max (a min y) = a min y = a min (x max y).
The second equation is handled in a similar manner.
If it is not a distributive lattice, then it has a sub-lattice isomorphic either to the diamond $M_3$ or the pentagon $N_5$.
It's easy to check that neither of these is distributive as a semi-lattice.
In the case of $M_3$, just take $x,y,z$ to be the atoms;
in the case of $N_5$, take, for example
$$0\prec x \prec z \prec 1 \quad\text{and}\quad 0 \prec y \prec 1.$$
While all of the above is true, it doesn't prove that a distributive semilattice which is also a lattice is a distributive lattice.
The reason is that the condition that defines distributive semilattices is not equational (or quasi-equational for that matter) and so, for the best of my knowledge, distributive semilattices don't form a class closed for the formation of subalgebras.
It turns out, it's not difficult to produce a direct proof.
So suppose that a lattice $L$ satisfies the condition
$$x\leq y \vee z \implies \exists y',z' \left( y'\leq y,\, z'\leq z,\, x=y'\vee z' \right).$$
Now we want to prove that $L$ satisfies
$$x\land (y\lor z)=(x\land y)\lor(x\land z).$$
Certainly $L$ (as all lattices) satisfy
$$x\land (y\lor z)\geq(x\land y)\lor(x\land z),$$
so we have to prove the converse.
But
$$x\land (y \lor z) \leq y \lor z,$$
and so there exist $y'\leq y$ and $z'\leq z$ such that
$$x\land(y\lor z)=y'\lor z'.$$
But then it follows that $x\geq y'\vee z'$, and so $x\geq y'$ and $x\geq z'$, yielding
$$y'=x\wedge y' \quad\text{and}\quad z'=x\wedge z'.$$
Hence
$$x\land(y\lor z)=(x\land y')\lor(x\land z')\leq (x\land y)\lor (x\land z).$$
Best Answer
We will construct a function from the natural numbers onto a distributive lattice, namely the lattice of all infinite sequences of natural numbers, all but finitely many of which are 0, with coordinatewise maximum as the join and coordinatewise minimum as the meet. The function is a lattice isomorphism. Let $p_1,p_2,\ldots$ be all prime numbers indexed in order. If $$n=p_1^{i_1}p_2^{i_2}\cdots$$ map $n\mapsto (i_1,i_2,i_3,\ldots)$. Then this is a lattice homomorphism, join corresponds to LCM and meet corresponds to GCD.
If you can't use this answer directly, consider this as a hint that LCM means you are taking the maximum of all powers of particular primes among the two numbers and GCD means you're taking the minimum.