Verification of Proof strategy I  am tasked with proving the following :
$$A \cap B^c \subseteq  (A \cap B)^c$$
I came up with the idea of using a combination of De Morgan's laws, rule simplification and rule of addition to prove this proposition:
$$A \cap B^c \subseteq  (A \cap B)^c \leftrightarrow  A \cap B^c \subseteq  A^c \cup B^c$$ 
However I am doubtful of the validity of my idea 
 A: $$\text{Goal: }\;A \cap B^c \subseteq  (A \cap B)^c$$
$$\begin{align} x \in (A\cap B^c) &\iff x \in A \land x\notin B\\ \\
&\implies x \in B^c \\ \\ & \implies x \in A^c \lor x\in B^c\\ \\ 
&\implies x \in (A^c\cup B^c)\\ \\ 
& \implies x \in (A\cap B)^c
\end{align}$$
A: $
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\calcop}[2]{\\ #1 \quad & \quad \text{"#2"} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
$Here are two alternative proofs that might give some more insight.
First, as you already did, we can transform the original statement using DeMorgan, to make both sides more similar:
$$\calc
A \cap B^c \subseteq  (A \cap B)^c
\calcop{\equiv}{set theory: DeMorgan on right hand side}
\tag{*} A \cap B^c \subseteq  A^c \cup B^c
\endcalc$$
Now a quick Venn diagram drawing suggests that $\text{(*)}$ is an instance of a more general theorem:
$$\calc
P \cap R \;\subseteq\; Q \cup R
\calcop{\equiv}{definitions of $\;\subseteq, \cap, \cup\;$}
\langle \forall x :: x \in P \land x \in R \;\Rightarrow\; x \in Q \lor x \in R \rangle
\calcop{\equiv}{logic: rewrite $\;p \Rightarrow q\;$ as $\;\lnot p \lor q\;$; DeMorgan on left hand side}
\langle \forall x :: x \not\in P \lor x \not\in R \lor x \in Q \lor x \in R \rangle
\calcop{\equiv}{logic: excluded middle for $\;x \in R\;$}
\langle \forall x :: x \not\in P \lor x \in Q \lor \text{true} \rangle
\calcop{\equiv}{logic: simplify}
\text{true}
\endcalc$$
This completes the proof.
As another alternative, you may recognize that the statement is of the form $\;P \subseteq Q^c\;$, and that that is the same as $\;P \cap Q = \emptyset\;$, which leads to the following proof:
$$\calc
A \cap B^c \subseteq  (A \cap B)^c
\calcop{\equiv}{set theory: the rule mentioned above}
A \cap B^c \cap A \cap B = \emptyset
\calcop{\equiv}{set theory: $\;B \cap B^c = \emptyset\;$}
A \cap \emptyset = \emptyset
\calcop{\equiv}{set theory}
\text{true}
\endcalc$$
A: $$A \cap B^c \subseteq  (A \cap B)^c
\iff (A \cap B^c)\cup  (A \cap B)^c
= (A \cap B)^c
$$
Just compute LHS:
$$
(A \cap B^c)\cup  (A \cap B)^c
= (A \cap B^c)\cup  (A^c \cup B^c)
= (A \cup  A^c \cup B^c) \cap (B^c \cup  A^c \cup B^c)  
=  \Omega \cap (B^c \cup  A^c )  = (B^c \cup  A^c ) =  (A \cap B)^c
$$
