NB: This is an alternative-proof question.
In this answer to . . .
Prove that $A \vartriangle B \subseteq C$ iff $A \cup C = B \cup C$.
. . . I mentioned to the OP that I believe the result can be shown using "iff" statements all the way. In the comments, I said I'd try that approach.
So far, I have
$$\begin{align} A\triangle B\subseteq C&\iff (A\triangle B)\cup C=C\\ &\iff ((A\setminus B)\cup(B\setminus A))\cup C=C\\ &\iff (\{a\in A\mid a\notin B\}\cup\{b\in B\mid b\notin A\})\cup C=C\\ &\iff \{c\in C\mid \color{red}{(c\in A\land c\notin B)}\\ &\lor\color{blue}{(c\in B\land c\notin A)}\}=C\\ &\iff \{c\in C\mid (\color{red}{(c\in A\land c\notin B)}\lor \color{blue}{(c\in B)})\\ &\land (\color{red}{(c\in A\land c\notin B)}\lor \color{blue}{(c\notin A)})\}=C\\ &\iff \{c\in C\mid ((\color{red}{(c\in A)}\lor\color{blue}{(c\in B)})\\ &\land(\color{red}{(c\notin B)}\lor\color{blue}{(c\in B)}))\\ &\land((\color{red}{(c\in A)}\lor\color{blue}{(c\notin A)})\\ &\land(\color{red}{(c\notin B)}\lor\color{blue}{(c\notin A)}))\}=C\\ &\iff\{c\in C\mid (\color{red}{(c\in A)}\lor\color{blue}{(c\in B)})\\ &\land (\color{red}{(c\notin B)}\lor\color{blue}{(c\notin A)})\}=C\\ &\iff C\cup((A\cup B)\cap(A^c\cup B^c))=C, \end{align}$$
but this leads me to $C\cup\varnothing =C$, which is vacuous.${}^\dagger$
So here's my question:
How does one prove "$(A\triangle B)\subseteq C$ iff $A\cup C=B\cup C$" using only "iff" statements?
This comment delineates an approach but I'd prefer something that starts out as my attempt does.
Please help :)
$\dagger$: Where did I go wrong?
$(A\cup B)\cap(A^\complement\cup B^\complement)$ contains anything that's in $A$ or $B$ but not both, so in general isn't $\emptyset$.