The first part of my question is:
Prove that if $\alpha$ or $\beta$ is a sentence, that $\mathfrak{A} \vDash (\alpha \vee \beta) \Rightarrow (\mathfrak{A} \vDash \alpha $ or $\mathfrak{A} \vDash \beta)$.
I can see how this would be easy to prove if $\alpha$ and $\beta$ we're both sentences from the definitions and the fact that if a sentence is satisfied with one variable assignment function, then it is satisfied for them all. But I cant seem to figure out how to do it in the case where only one of $\alpha$ or $\beta$ is a sentence?
Then my other question would be, how/why this can fail when $\alpha$ and $\beta$ are not sentences? Does it rely on certain aspects of the language? For instance if I had a very simple language with just one relation symbol, what would a counter example be, or would one even exist?
The issue is with the formal definition of satisfaction of a formula :
If $\varphi$ is a sentence, then either $\mathfrak A$ satisfies $\varphi$ with every function $s$ from $\text {Var}$ into $|\mathfrak A|$, or $\mathfrak A$ does not satisfy $\varphi$ with any such function.
What is the definition of $\mathfrak A \vDash \alpha$ when $\alpha$ has a free variable $x$ ? We may adopt the convention that $\mathfrak A \vDash \alpha$ iff $\mathfrak A \vDash \forall x \alpha$.
In this case, the question amounts to :
If neither $\alpha$ nor $\beta$ are sentences, the property does not hold : consider the structure $\mathbb N$ and the formula $\forall x (\text {Even}(x) \lor \text {Odd}(x))$.
Let assume that $\alpha$ is a sentence and consider $\mathfrak A$ whatever such that $\mathfrak{A} \vDash \alpha \vee \beta$, i.e. $\mathfrak{A} \vDash \forall x \ (\alpha \vee \beta)$.
If $\alpha$ holds in $\mathfrak A$, it's done.
So, consider the case : $\mathfrak{A} \nvDash \alpha$. This means that $\alpha \lor \beta$ must be satisfied by every $s$, and due to the fact that $s$ does not satisfies $\alpha$, we have $\mathfrak A \vDash \beta[s]$, for every $s$.