A Simple SAT Instance:
Let $R \supseteq \{p_1, p_2, p_3, p_4, p_5\}$.
Let $F = (¬p_1 ∨ p_2) ∧(¬p_2 ∨ p_1) ∧ (¬p_1 ∨ ¬p_2 ∨ ¬p_3) ∧ (p_1 ∨ p_2) ∧ (¬p_4 ∨ p_3) ∧ (¬p_5 ∨ p_3)$.
$\left\{p_1, p_2\right\}$ is a model for $F$.
Hence, $F$ is satisfiable.
This is a simple example of a SAT Instance. I understand what it means, and how to get the result, but I don't understand why only $\left\{p_1, p_2\right\}$ is a model for $F$.
If you consider $(\neg p_1 \vee p_2) \wedge (p_1 \vee p_2)$ together, they tell you that $p_2$ must be true. The second clause, $(\neg p_2 \vee p_1)$, then tells you that $p_1$ must be true. Substitute those truth values for $p_1$ and $p_2$ in the third clause. The third clause then implies $p_3$ must be false, and finally the last two clauses imply that $p_4$ and $p_5$ must be false.
We only found necessary consequences of $F$. Hence the values we found give the only solution.
Evidently, a model in your context is a subset $M$ of $R$ such that assigning value true to all elements of $M$ and value false to all elements of $R \setminus M$, $F$ is satisfied.