Application of compactness theorem. A partial order $<$ on a set $A$ has dimension less than n+1 if there exists a sequence of n linear orders {$<_1,.....,<_n$} on $A$ such that:
$\forall\ x,\ y\in A, x<y$ iff $(x<_1y\ \land .......\land\ x<_ny)$
Show that a partial order on a set A has dimension less than $n+1$ iff for every finite subset $X\subseteq A$, the restriction of $<$ on $X$ has dimension less than $n+1$.
This question is practically begging me to use compactness just by it's phrasing, but the problems is I have NEVER use compactness before in my life, hence I'm stuck after a few steps. Below is my attempt.
Compactness theorem: A set of wffs is satisfiable iff every finite subset is satisfiable
Attempt:
Pick any finite subset $X\subseteq A$. Let $<$ be restricted to X and has dimension less than $n+1$
Fix an $x, y$.
Let $<$ be any linear order.
Define  truth assignment $v$ as follow,
$v($<$):= \begin{cases}
T\ \iff x<y\\  
F \ otherwise
\end{cases}$
So now I want to mould the assumption I made so that I can say X is satisfiable, but  I dont exactly know what it means to be satisfiable in this context.
Any help or insights is deeply appreciated.
To any veteren of using compactness, it would mean so much if you could share when and how to use compactness in general
 A: We can indeed use compactness here! Although it's much easier to use the version for first-order logic than the propositional one, so that's what I'll do.
One direction is trivial. Now suppose for every finite set $S$, there are linear orders $<_i^S$ with the desired property.
Then consider the following language $L$:


*

*We have a binary relation symbol "$<$"

*. . . and $n$ more binary relation symbols "$<_i$" $1\le i\le n$

*. . . and a bunch of constant symbols, $c_a$, for $a\in A$;
and the theory $T$ in this language consisting of the following sentences:


*

*$<$ is a partial order and $<_i$ is a linear order for each $1\le i\le n$

*the sentence "$c_a<c_b$" for $a<b$

*and the sentence "$\forall x, y(x<y\iff [(x<_1y)\wedge (x<_2y)\wedge . . . \wedge (x<_3y)])$".
By assumption, every finite subset of $T$ has a model; by Compactness, the whole of $T$ has a model. Now let $M$ be any such model; we get some binary relations $\prec_i$ on $A$ for $1\le i\le n$, given by $$a\prec_ib \iff M\models c_a<_ic_b.$$ And it's not hard to show that these relations have the desired property.

OK, now what about propositional logic? Can we make do with just the propositional version of compactness? 
The answer is yes, but it's somewhat ad hoc. Here's what we do. We consider, as above, a really big language. It's a propositional language now, so it just consists of "propositional atoms" (or letters, or whatever your text calls them). Here are the ones we'll use (and these are all distinct from each other):


*

*For $a, b\in A$ and $1\le i\le n$, we have the atom "$p_{a, b, i}$", which intuitively means "$a<_ib$".

*For $a, b\in A$ we have the atom "$q_{a, b}$", which intuitively means "$a<b$".
Now, can you see how to write down a set $S$ of propositions in these symbols which expresses "the relations $<_1, <_2, . . . , <_n$ show that $A$ is $<(n+1)$-dimensional"?
Do you see why - assuming each finite subset of $A$ is $<(n+1)$-dimensional - $S$ is finitely satisfiable?
Finally, do you see how to go from a truth assignment for $S$ to the desired family of orderings on $A$?
