Proof by Elements to Show $D^{c} ⊆ A^{c}$ 
Use proof by elements to verify that for all nonempty sets $A$, $B$, and $D$ if $A ⊆ B$, $D^{c} ⊆ B^{c}$, then $D^{c} ⊆ A^{c}$.

Here's the proof I have written so far. I have gotten feedback that I am missing steps and have used incorrect algebra, but I can't see where the error in my logic is:
Proof by elements:
$\forall x, x∈D^{c} → x∈B^{c}$ since $D^{c} ⊆ B^{c}$
$→ x∈A^{c}$ since $A ⊆ B \iff B^{c} ⊆ A^{c}$
So, by definition of containment, $D^{c} ⊆ A^{c}$, $\forall x$.
Any help or guidance in the right direction would be greatly appreciated!

I'd use less symbols and more words.
The aim is to show that, if $x\in D^c$, then $x\in A^c$.
So, let $x\in D^c$. By hypothesis, $x\in B^c$, that is, $x\notin B$. Since $A\subseteq B$, we also have $x\notin A$ (otherwise $x\in B$). So $x\in A^c$.
Your proof is good, anyway.
