Caution: The existence of an injection between $A$ and $C$ doesn't necessarily imply $A \subseteq C$.
For example, consider set $A$, the set of all even integers, and set $B$, the set of all odd integers. Certainly, there exists an injection: $f: A \to C$, $\;f(x) = x + 1\;$ (which is not only injective, but surjective, as well). But clearly, $\;A \nsubseteq C$.
The converse is true: if $A \subseteq C$, then an injection $h: A \to C$ exists.
But we can easily prove the inclusion $A \subseteq C$ by "element chasing:" a standard way to prove set inclusions, and/or set equivalencies.
We have $A \subseteq B$ and $ B \subseteq C$. And we want to prove that this necessarily implies $A\subseteq C$.
$(1)$ Suppose $x \in A\quad $ (Assumption)
$(2)$ We know $A \subseteq B$ means $x \in A \implies x \in B.\;$ So given $(1)$, we have $x \in B$.
- $(3)$ We know $B \subseteq C$ means $x \in B \implies x \in C$. So given $(2)$, we have $x \in C$.
$(4)\;\;x \in A \implies x \in C$. $\quad[(1) - (3)]$
Therefore, $A \subseteq C$.