If $A$ is a subset of $B$, then the closure of $A$ is contained in the closure of $B$.

I think it's much simpler than that. By definition #1, the closure of A is a subset of any closed set containing A; and the closure of B is certainly a closed set containing A (because it contains B, which contains A). QED.

Using Definition #1 makes it quite easy. For each $A \subseteq X$, let $\mathcal{C}_A = \{ F \subseteq X : F\text{ is closed and }A \subseteq F \}$. Then by Definition #1 it follows that $\overline{A} = \bigcap \mathcal{C}_A$.

Note that if $A \subseteq B$, then $\mathcal{C}_B \subseteq \mathcal{C}_A$, and therefore $\bigcap \mathcal{C}_A \subseteq \bigcap \mathcal{C}_B$.

I think it's simplest to see from the first definition.

Let $\mathcal{A}$ be the collection of closed sets containing $A$ and $\mathcal{B}$ the collection of closed sets containing $B$. Since $A \subset B$, we know $\mathcal{B} \subset \mathcal{A}$, and so $\bigcap \mathcal{A} \subset \bigcap \mathcal{B}$ (i.e. $\overline{A} \subset \overline{B}$).

Loosely speaking, adding more sets to an intersection can only make it smaller.

You say that some of the proofs you have looked use the argument "that since $A$ is contained in $\overline{B}$, then $\overline{A}\subseteq\overline{B}$" and that they don't seem strong enough for you but this follows directly from definition #1. Any closed subset containing $B$ contains $A$ and consequently $A\subseteq \overline{B}$. Since $\overline{B}$ is closed, $\overline{A}\subseteq\overline{B}.$