In the following formula $\alpha \rightarrow \forall x \alpha$, why is it important that $x$ does not occur free in $\alpha$?
Here are my thoughts: I would like to find a countermodel and formula $\alpha$ where $x$ is free in $\alpha$ and $\alpha \rightarrow \forall x \alpha$ holds. If we take $\alpha$ to be the formula $x = 5$ in some model whose domain is $\mathbb{N}$, we see that neither $\alpha$ nor $\forall x \alpha $ is satisfied which is not what we want. I cannot see how to meaningfully satisfy $\alpha$ while not satisfying $\forall x \alpha$. Any hints?
Using a semantic approach to classical first-order logic, we usually interpret formulas as subsets of some power of the domain. That is, a formula $\varphi$ with $n$ free variables is interpreted as a subset $[\![\varphi]\!]\subseteq D^n$ where $D$ is the domain.
Skipping some details and being a bit sloppy, $\varphi\to\psi$ is valid if $[\![\varphi]\!]\subseteq[\![\psi]\!]$. Assuming $\alpha$ has no free variables except $x$ for simplicity, then $[\![\forall x.\alpha]\!]$ is either $\varnothing$ if $[\![\alpha]\!]\neq D$ or it is $D$ if we view $\forall x.\alpha$ as being a formula which potentially but not actually contains $x$ free. Thus, if $\alpha$ is holds for anything, i.e. $[\![\alpha]\!]\neq\varnothing$, but doesn't hold for everything, i.e. $[\![\alpha]\!]\neq D$, then $[\![\alpha]\!]\not\subseteq[\![\forall x.\alpha]\!]$ as this is $[\![\alpha]\!]\not\subseteq\varnothing$ in this case.
So your example works fine as a counter-example. You can arrange things in a way that is arguably a bit more clearer if you ask instead about $\alpha\vDash\forall x.\alpha$.