I want to show why Modal Logic tries to incorporate the two following implications
$\square A \to \square \square A$
$\diamond A \to \square \diamond A$
as valid sentences by showing that a natural understanding of Necessity and Possibility leads to those two implications as valid and since Modal Logic wants to mimic our natural understanding of Necessity and Possibility as closely as possible it is bound to incorporate those two implications as valid sentences (theorems). So I wanna prove these implications with naive/informal Modal Logic. I wanna know if I succeed with my proofs.
Proof $\square A \to \square \square A$
Assume $\square A$. Now in addition we hypothetically assume $\neg \square \square A$, i.e. $\diamond \neg \square A$, i.e. one can think of at least one instance where $\neg \square A$ but since in that very instance our assumption $\square A$ holds as well we got a contradiction, and so we can infer $\square \square A$.
Proof $\diamond A \to \square \diamond A$
Assume $\diamond A$. Now in addition we hypothetically assume $\neg \square \diamond A$, i.e. $\diamond \neg\diamond A$, i.e. one can think of at least one instance where $\neg\diamond A$ but since in that very instance our assumption $\diamond A$ holds as well we got a contradiction, and so we can infer $\square \diamond A$.
These statements are not true in general for modal logics.
For $\square A \to \square\square A$, consider the following frame:
The worlds are $1, 2, 3$ and $R = \{(1, 2), (2, 3)\}$.
Suppose $A$ holds precisely at world $2$.
Then, starting from world $1$, $\square A$ hold but $\square\square A$ fails.
For $\lozenge A \to \square \lozenge A$, the example above works as well.
$\lozenge A$ holds of world 1, but $\square \lozenge A$ does not.
These two formulas are actually imposing conditions on the accessibility relation in our frame.
$\square A \to \square \square A$, if $A$ is allowed to range over arbitrary variables (or arbitrary formulas) and we start from any world in the frame, insists that the accessibility relation is transitive:
$$ aRb \land bRc \implies aRc \;\; \text{for all worlds $a,b,c$} $$
I think this is easier to see if we note that $\square A \to \square\square A$ expresses the same property as $\lozenge\lozenge A \to \lozenge A$.
$$ [\forall A](\square A \to \square\square A) \iff [\forall A](\square \lnot A \to \square\square \not A) \iff [\forall A](\lnot \lozenge A \to \lnot \lozenge \lozenge A) \iff [\forall A](\lozenge \lozenge A \to \lozenge A) $$
Intuitively, this says that any world we can reach in two jumps can be reached in one jump, which is transitivity.
Similarly, $\lozenge A \to \square\lozenge A$ expresses the idea that any reachable world is reachable remains reachable from all subsequent worlds.
$$ aRb \land aRc \to bRc $$
This is a Euclidean relation, which is kind of an obscure condition that I've only ever seen in modal logic.
Interestingly, a relation being transitive and Euclidean does not imply that it's an equivalence relation.
The empty relation on any set is trivially not an equivalence relation, but also ...
Consider the following relation on $\{1, 2\}$.
$$ \{ (1, 2), (2, 2) \} $$
And the following relation on $\{1, 2, 3\}$.
$$ \{ (1, 2), (1, 3), (2, 2), (3, 3), (2, 3), (3, 2) \} $$