How should I go about doing this proof? I am new to mathematical proofs and would like some help understanding how to prove 
$$
\left(A^c \cup B^c\right) - A = A^c
$$
I would like to see a proof if possible. I understand that we need to prove equality of the two sides which is done by making sure both sides are subsets of each other.
 A: $\textbf{gt6989b}$'s method is probably the one you'll want to use for a beginner proof. Alternatively, you could do this with some set algebra. We can use the identity $$C \setminus D = C \cap D^c$$ to rewrite your equation as $$\left(A^c \cup B^c\right) - A  = \left(A^c \cup B^c\right) \cap A^c $$ Can you proceed from here using the distributive law of set algebra?
A: hints
Every time you need to prove equality of two sets, $A=B$, you must show


*

*If $a \in A$, then $a \in B$.

*If $b \in B$, then $b \in A$.


In your case, the first part starts out like this this. Let $x \in \left(A^c \cup B^c\right) - A$. Now prove $x \not \in A$.
Then, let $y \not \in A$ and prove that $y \in \left(A^c \cup B^c\right) - A$.
UPDATE
Note that since $x \in \left(A^c \cup B^c\right) - A$, and the last operation ($-A$) removes all elements of $A$ from consideration, what can be said about $x$?
Update 2 Can you prove that if $x \in Z - A$ for any set $Z$, then $x \not \in A$? This will finish part 1.
Now for part 2, let $y \in A^c$. Thus, $y \in A^c \cup B^c$ (why?), and since $y \not \in A$, removing $A$ from the set does not affect the presence of $y$, hence, $y \in \left(A^c \cup B^c\right) - A$ as desired.
A: Trivially $\left(A^c \cup B^c\right) - A\subseteq U-A = A^c$; (because $\left(A^c \cup B^c\right) \subseteq U$, where $U$ is universal set). 

For Proof of  $ A^c\subseteq\left(A^c \cup B^c\right) $, we can use two facts:
fact. 1- By De Morgan's laws, $\left(A^c \cup B^c\right)=\left(A\cap B\right)^c$.
fact. 2- If $T \subseteq S$ then $S^c \subseteq  T^c$.
Now
Let $\color{red}{x\in A^c}$. By "fact 2", $x\in \left(A\cap B\right)^c$. By "fact 1", $\color{red}{x\in \left(A^c \cup B^c\right)}$. So $x\in \left(A^c \cup B^c\right)-A$ 
