Is this proof of $(S \cap T)^c=S^c \cup T^c$ valid? I'm self-learning set theory, and as an exercise, I tried to prove one of DeMorgan's laws, i.e., 

$$(S \cap T)^c=S^c \cup T^c$$

So my proof goes as follows:
Let $x \in S^c$, and $x \in T^c$. Then, $x \notin S$ and $x \notin T$. Therefore, $x$ is neither a member of $S \cup T$, nor a member of $S \cap T$. As a result, it can be seen that $(S \cap T)^c \subset S^c \cup T^c$. Being that $x$ is a member of both $(S \cap T)^c$ and $S^c \cup T^c$, it follows that $S^c \cup T^c \subset (S \cap T)^c$. Therefore, being that $(A=B) \iff (A\subset B) \land (B\subset A)$, we can conclude that $(S \cap T)^c=S^c \cup T^c$. QED. 
My question is, does this prove DeMorgan's law?
 A: Good question; however, I do not have much time to answer so I hope you can forgive me for being a little blunt. 
The proof is confusing at best. 
First, it's unclear why you start with "Let $x \in S^c$, and $x \in T^c$. You are taking an $x$ from $T^c \cap S^c$, and it's unclear why, because this set does not appear in the original statement. You should be taking an $x$ from $(S \cap T)^c$ and try to show it's in $S^c \cup T^c$, and vice versa.
Then you write "As a result, it can be seen that $(S∩T)^c \subset S^c ∪T^c$". Maybe you are able to see this, but the proof so far has not communicated why that is the case; the $x$ you took had nothing to do with the sets $(S∩T)^c$ and $S^c ∪T^c$, since again, the $x$ you took is in  $T^c \cap S^c$.
Half of a proof (I leave the other half to you) is:
Take any $x \in (S \cap T)^c$. Then $x \not \in S \cap T$, so $x \not \in S$ or $x \not \in T$. Therefore $x \in T^c$ or $x \in S^c$, so $x \in T^c \cup S^c$.
Feel free to ask any question and I will answer when I can.
