Just how are you confused about drawing the lattice? Here's how I did it. I started by putting 1 at the bottom, since it's the least element of the lattice, it divides everything.
Since 2 and 3 are divisible by 1, but not by each other, I put 2 one cm northwest of 1, I put 3 one cm northeast of 1, and draw lines from 2 to 1 and from 3 to 1.
Since 6 is divisible by everything so far, I put 6 one cm NE of 2 (and one cm NW of 3) and draw lines from 6 to 2 and from 6 to 3. Of course I don't draw a line from 6 to 1.
I now have a diamond, with 1 at the bottom, 2 and 3 on the sides, 6 at the top.
Next I look at 4: divisible by 1 and 2, not divisible by 3 or 6. So I put 4 one cm NW of 2, and draw a line from 4 to 2.
12 is divisible by everything so far. I put 12 one cm NE of 4 (and NW of 6), and draw lines from 12 to 4 and from 12 to 6.
9 is divisible by 1 and 3, but not divisible by 2, 4, 6, or 12. I put 9 one cm NE of 3, and draw a line from 9 to 3.
18 is divisible by 6 and 9, and therefore also by 1, 2, and 3; but 18 is not divisible by 4 or 12. I put 18 one cm NE of 6 (NW of 9) and draw lines from 18 to 6 and 9.
That's the diagram. I'd draw it for you if I knew how.
Looking at the diagram, it's easy to see that it's a lower semilattice, meaning that any two elements have a greatest lower bound.
It's also easy to see that it's not an upper semilattice (and therefore not a lattice); the pairs {4,9}, {4,18}, {9,12}, and {12,18} do not have least upper bounds.
By the way, what did you think was the least upper bound of 12 and 18? Maybe there was a typo in your question, and 36 was supposed to be an element of S?
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.
Best Answer
To show that a partial order is a lattice, you have to show that it has well-defined "meet" and "join" operations. Because the partial order here is $\subseteq$ and the underlying set is the full powerset, meet and join are $\cap$ and $\cup$ respectively.
To show that a set of sets $S$ is a lattice with respect to $\subseteq$, you have to show that it's closed under those operations. When "meet" and "join" are actual intersection and union, this amounts to showing: $\forall X, Y \in S$, both $X \cap Y \in S$ and $X \cup Y \in S$. When $S = \mathcal{P}(B) = 2^B$ is the powerset of a set $B$, there isn't a lot to actually "prove". You have to show that $X \cap Y$ is actually the largest set contained ($\subseteq$) in both $X$ and $Y$, and that $X \cup Y$ is actually the smallest set containing ($\subseteq$) in both $X$ and $Y$... and that both are actually subsets of $B$ :)
Note, however, that even when the underlying set is a set of sets, and the partial order is $\subseteq$, "meet' and 'join' may not actually be set-theoretic $\cap$ and $\cup$. Example: a four-element diamond lattice $\{\emptyset, \{0, 1\}, evenInts, \mathbb{N}\}$, where $evenInts$ is... the set of even integers. The "meet" of $\{0, 1\}$ and $evenInts$ is the bottom of the diamond, $\emptyset$, but of course their intersection is nonempty; similarly, the "join" of $\{0, 1\}$ and $evenInts$ is the top of the diamond, $\mathbb{N}$, but their union doesn't equal $\mathbb{N}$.