Why is the type ∀t.t un-inhabited in System F?
The simplest proof is giving a model where types are interpreted as propositions, and terms as proofs, then observing that $\forall \alpha.\alpha$ is interpreted as the false proposition, so any $\cdot \vdash t : \forall \alpha.\alpha$ would be a proof of falsehood. It is important to remember that $\forall \alpha.\alpha$ is only uninhabited for sure in the empty context $\cdot$, in other contexts it may be trivially inhabited.

Concretely, first interpret each $A$ type with $n$ free type variables in the following way:

$$
\begin{alignat*}{2}
&amp; [\![ A ]\!] &amp;&amp; : \mathsf{Bool}^n \to \mathsf{Bool}\\
&amp; [\![ \alpha ]\!]\,\gamma &amp;&amp; := \text{$\alpha$-th component of $\gamma$}\\
&amp; [\![ A \to B ]\!]\,\gamma &amp;&amp; := [\![ A ]\!]\,\gamma \Rightarrow [\![ B ]\!]\,\gamma\\
&amp; [\![ \forall \alpha.\,B]\!]\,\gamma &amp;&amp; := [\![B]\!]\,(\mathsf{true},\,\gamma)\,\land\,[\![B]\!]\,(\mathsf{false},\,\gamma)
\end{alignat*}
$$

Interpret every typing context $\Gamma$ with $n$ free type variables as:

$$
\begin{alignat*}{2}
  &amp; [\![\Gamma]\!] : \mathsf{Bool}^n\to \mathsf{Bool}\\
  &amp; [\![\Gamma]\!]\,\gamma := \bigwedge\limits_{A\,\in\,\Gamma}\,[\![A]\!]\,\gamma
\end{alignat*}
$$

Then show for each $\Gamma \vdash t : A$ term, that for each truth valuation $\gamma : \mathsf{Bool}^n$, if $[\![\Gamma]\!]\,\gamma = \mathsf{true}$, then $[\![A]\!]\,\gamma = \mathsf{true}$. This can be done by induction on terms.

Now, $[\![\forall \alpha.\alpha]\!]$ with no free type variables evaluates to $\mathsf{false}$, since it's $\mathsf{true}$ when $\alpha$ is instantiated to $\mathsf{true}$, and $\mathsf{false}$ when it's instantiated to $\mathsf{false}$, and the conjunction of these is $\mathsf{false}$. Assuming $\cdot\vdash t : \forall \alpha.\alpha$, we have by the previous result that $[\![\forall \alpha.\alpha]\!]$ is $\mathsf{true}$, a contradiction, hence there is no $\cdot\vdash t : \forall \alpha.\alpha$.