Can you form a sentence in $\mathcal{L}=\{\leq\}$ and structure $\mathcal{N}=\{\mathbb{N},\leq\}$ which states for any $x$ not $0$ there are only finitely many elements, less then or equal to $x$?
The definition of sentence I have is the same as this wikipedia article, so I'm assuming it's a standard definition, that being a well formed formula with no free variables.
I know the definition of a formula does not allow for infinite many terms in a formula. So my assumption is that you cannot form a single sentence which says this. But could you form a countable collection of sentences which would be this property?
No, you cannot do this.
(And countability doesn't come into play: there is no set $\Gamma$ of first-order sentences such that $\Gamma$ is true in exactly those structures with the countable predecessor property. Now in fact in our case since the language is finite all sets of sentences in the language are countable a priori, but it's still worth observing that the following argument style is quite "coarse.")
First, let's see why it's fairly implausible.
The natural way to say "every element has only finitely many predecessors" uses an infinite disjunction inside a universal quantifier: $$\forall x[\bigvee_{n\in\mathbb{N}}\exists y_1,...,y_n\forall z(z\le x\rightarrow z=y_1\vee ...\vee z=y_n)].$$
Now, if that "$\bigvee$" were a "$\bigwedge$" we could turn this into a countably infinite conjunction of first-order sentences since $$\forall x\bigwedge_{n\in\mathbb{N}}\psi_n(x)\mbox{ is equivalent to }\bigwedge_{n\in\mathbb{N}}\forall x\psi_n(x).$$ But that trick doesn't work here. First of all, "$\forall$" and "$\bigvee$" don't commute; but even if they did, the "$\bigvee$" is fatal since it makes us want an infinite disjunction and a countable set of sentences corresponds to an infinite conjunction.
Of course, the above doesn't prove that you can't do this; it's merely a heuristic argument.
To prove that a property is not first-order expressible, there are two main method: compactness and downward Lowenheim-Skolem (upward Lowenheim-Skolem is a corollary of compactness). The following is a decent heuristic (but as always don't push it too far): we choose which one to apply based on whether the "bad model" we want to show exists is bad because it's too big (in which case we use compactness) or too small (in which case we use DLS).
In this case, we want to show that if $\varphi$ is a sentence true in every structure where each element has only finitely many predecessors then it "overspills" - that is, there is some structure where some element doesn't have only finitely many predecessors but where $\varphi$ still holds. This tells us that we want to apply compactness.
So: do you see how to use compactness to show that if $\varphi$ is true in every structure with the finite predecessor property, it's also true in some structure without the finite predecessor property? (HINT: it's going to look almost identical to the proof that there's no first-order sentence true in exactly the finite structures. Remember, the key trick there was expanding the langauge ...)