discrete mathematics
I need to prove that if $R,S$ are both partially ordered sets (POsets)
then $R \cap S$ is a POsets as well.
I got stuck while proving it is transitive:
let $<a,b>,<b,c> \in R\cap S$ then, WLOG, $<a,b>,<b,c> \in R$
and thus $<a,c> \in R$ because R is a POSet.
Similarly $<a,c> \in S$ and thus $<a,c> \in R \cap S$
Is it considered a proof of transitivity? Are there any mistakes along the way?
The existence of any counterexample is a proof that the statement is false. The statement is of the form
"For all transitive relations, if $S$ and $R$ are transitive relations, then $S\cup R$ is transitive." $\quad (*)$
Hence, to prove this statement is false, it suffices to prove the existence of two transitive relations whose union is not transitive. This is what a counterexample proves.
That is, the negation of $(*)$ is
$\lnot (*)$ "It is NOT THE CASE that for all transitive relations, S, R, that $S \cup R$ is transitive."
which is equivalent to
$\lnot (*)$"There exists transitive relations $S, R$ such that $S \cup R$ is NOT transitive."
The whole idea of a Hasse diagram is just an efficient representation of your poset. If you think about the set of subsets of $\{1,2,3\}$ ordered by inclusion (that is, $\subseteq$), we have ordered pairs like $(\varnothing, \{1\}),\, (\varnothing \{1,2\}),\, (\{1\}, \{1,2\})$, and so on, reflecting the fact that $\varnothing \subseteq \{1\}$, etc.
But since we know that $\subseteq$ is transitive, then knowing $\varnothing \subseteq \{1\}$ and $\{1\} \subseteq \{1,2\}$ tells us that $\varnothing \subseteq \{1,2\}$; it would be silly for our efficient representation to waste time representing this, as long as we make sure to represent $\varnothing \subseteq \{1\}$ and $\{1\} \subseteq \{1,2\}$:
So for example, the fact that we can trace a path from $\{1\}$ up to $\{1, 2, 3\}$ using edges in the Hasse diagram means that $\{1\} \subseteq \{1,2,3\}$, or equivalently, that the ordered pair $(\{1\}, \{1,2,3\})$ is in our relation.
We only draw edges in the Hasse diagram to depict so-called covering relations: So, we drew a line from $\{1\}$ to $\{1,2\}$ because if we have an inequality like
$$\{1\} \subseteq \Delta \subseteq \{1,2\},$$ then we're forced to use $\Delta = \{1\}$ or $\Delta = \{1,2\}$; nothing "fits" between them properly. Thus, we say that $\{1\}$ is covered by $\{1, 2\}$, and draw a line in the Hasse diagram.
We generally use the symbol $\le$ any time we have a poset, even if the relation has nothing to do with the normal definition of $\le$ to compare numbers. So we could write $\varnothing \le \{1\}$ in the example above, if we wanted to. Some people, to prevent this confusion, use the symbol $\preceq$ for a generic poset, to help you realize that the relation might have nothing to do with the usual way to order real numbers.
When you're given an unlabeled Hasse Diagram as in your last example, just call all of the nodes by some name; $\{1, 2, 3,4\}$ or $\{a, b, c, d\}$, it doesn't really matter. Just that we can see the comparisons between them.
So we could pick 
and start getting ordered pairs $(a, c),\, (a, d),$ etc, since we can see that $a \le c$, $a \le d$, and so on. In this example, the only relations are covering relations, so we'll have as many ordered pairs as there are lines in the graph.
In your linear example with $1 \le 2 \le 3 \le 4$, we would have $4 + 3 + 2 + 1$ ordered pairs: $\{(i, j): i \le j\}$ even though there are only $3$ edges in the Hasse Diagram.
Best Answer
I think you mean that $R$ and $S$ are both partial orders on some set $X$, so that $X$ is the partially ordered set, not $R$ and $S$ themselves.
It’s fine, apart from one slight misuse of language. You don’t actually mean ‘without loss of generality’, since you’re not picking just one of the alternatives and saying that it doesn’t matter which one is picked: you’re picking both and observing that $\langle a,b\rangle$ and $\langle b,c\rangle$ are both in both $R$ and $S$. Thus, you can apply transitivity of both $R$ and $S$ to conclude, as you did, that $\langle a,c\rangle$ is in both $R$ and $S$ and hence in their intersection. Drop the WLOG, and you’re fine.