Proof over subsets So I'm currently taking a course for proofs, could you please check my work?
Prove if $B \subseteq C$ then $A \cup C^c$ is a subset of $A \cup B^c$.
For all $x \in B$, $x$ will be an element of $C$. So, by definition of complement, if there exists $y \notin C$, $y$ will be an element of $C^c$. Because $B$ is subset of $C$, $y$ will not be an element of $B$, but $y$ will be an element of $C^c$. Because the set $B^c$ contains all elements of $C$ that are not in $B$, $B^c$ will contain $C^c$. Thus $C^c$ will be a subset of $B^c$. By the definition of union sets $A \cup C^c$ will be a subset of $A \cup B^c$, thus proving our original claim. 
If you guys have any feedback on a way to write this please let me know. 
Thanks!
 A: About your proof, what is not so easy to grasp is the inference from :

$B \subseteq C$

to :

$C^C \subseteq B^C$.

We have, by definition of "subset" :

for all $x$, if $x \in B$, then $x \in C$.

Up to now it's Ok.
Then we consider $y \in C^C$; this means that $y \notin C$, by definition of "complement".
Because $y \notin C$, we must have that $y \notin B$. Assume the contrary, i.e. $y \in B$. By the previous result, then $y \in C$, and this is contradictory, because in this case we have both $y \notin C$ and $y \in C$.
Having proved that $y \notin B$, we have that $y \in B^C$.
But $y$ is "generic"; thus, we have that :

for all $y$, if $x \in C^C$, then $y \in B^C$,

i.e.


$C^C \subseteq B^C$.


Note. To be precise, it is not necessary to assume that "there exists $y \notin C$"; if not, $C^C$ is empty, thus $C^C = \emptyset \subseteq B^C$, because $\emptyset$ is subset of every set.
Now we can proceed with the last part of your proof :

Thus $C^C$ will be a subset of $B^C$. By the definition of union of sets $A \cup C^C$ will be a subset of $A \cup B^C$, thus proving our claim.

A: Let $x \in A \cup C^c$. If $x\in A$ then $x \in A \cup B^c$. If not then $x \in C^c$, but $B \subseteq C$ means $C^c \subseteq B^c$. Therefore $x \in C^c \subseteq B^c \subseteq A \cup B^c$.
