This has already been asked here, but I used an argument via contradiction as it seemed fitting in the context of the full proposition that needed to be proven.
Definition $3.1.14$ (Subsets). Let $A$, $B$ be sets. We say that $A$ is a subset of $B$, denoted $A ⊆ B$, iff every element of $A$ is also an element of $B$, i.e., For any object $x$, $x ∈ A \implies x ∈ B$.
We say that $A$ is a proper subset of $B$, denoted $A \subsetneq B$, iff $A ⊆ B$ and $A \neq B$.
This is from Tao's Analysis I. The definition of a proper subset has an "if" instead of the "iff" but I used the latter because of the errata.
The proposition in full says:
Proposition $3.1.17$ (Sets are partially ordered by set inclusion). Let $A$, $B$, $C$ be sets. If $A ⊆ B$ and $B ⊆ C$ then A$ ⊆ C$. If $A ⊆ B$ and $B ⊆ A$, then $A = B$. Finally, if $A \subsetneq B$ and $B \subsetneq C$ then $A \subsetneq C$.
Having already proven these two previous results, my proof went as follows.
Proof. Assume $A\subsetneq B$ and $B \subsetneq C$. By definition, $A \subsetneq B \implies (A \subseteq B)\land (A\neq B)$. Similarly, $B \subsetneq C \implies (B \subseteq C)\land (B\neq C)$. By the proposition we already know that $A \subseteq C$ so we only need to prove that $A\neq C$ to get the full result. Suppose for the sake of contradiction that $A=C$. Then, every element of $C$ is also an element of $A$, i.e., $C\subseteq A$. Also, $A \subseteq B$ so by proposition, $C \subseteq B$. So we have both $C\subseteq B$ and $C \subseteq B$, then by proposition, $B=C$, a contradiction. So $A\neq C$, which together with $A\subseteq C$ closes the proof.
Is this correct? Do you think this is the solution Tao had in mind?
The last proposition has a simple proof: just choose $b \in B$ such that $b \notin A$. By assumption $b \in C$ and we get $A \neq C$. This holds even when $B=C$ so that the condition $B \neq C$ is unnecessary.