I'm trying to prove $ M \subset N \Rightarrow M \cup(N \setminus M) = N $ I'm trying to prove:
$$
M \subset N \Rightarrow M \cup(N \setminus M) = N
$$
My trial:
$
\{x|x\in M \cup(N \setminus M)\} \Leftrightarrow \{x|x\notin \overline{M \cup(N \setminus M)}\}
$
using de Morgan
$ \Leftrightarrow \{x|x\notin \overline{M} \cap \overline{(N \setminus M)}\}$
$ \Leftrightarrow \{x|x\notin \overline{M} \wedge x \notin \overline{(N \setminus M)}\}$
$ \Leftrightarrow \{x|x\notin \overline{M} \wedge x \in (N \setminus M)\}$
$ \Leftrightarrow \{x|x\notin \overline{M} \wedge x \in N \wedge x\notin M\}$
aand we got a problem here. $x$ cannot be both $x\notin \overline{M}$ and  $x\notin M$.
My steps seems logical to me. Can someone please spot my mistake. Thanks a lot!
Your error: $x\notin A\cap B$ is the negation of $x\in A\cap B$. $$x\in A\cap B \text{ iff }x\in A \land x\in B$$
So,$$x\notin A\cap B \text{ iff }x\notin A \lor x\notin B$$
Please notice how the $\land$ becomes $\lor$ on negation! In general, $\lnot(p\land q) = \lnot p\lor \lnot q$.

Alternative proof strategy: Whenever I want to show $A = B$, where $A,B$ are two sets, I aim to establish $A\subset B$ and $B\subset A$. Works on most occasions!

Suppose $M\subset N$. Then, for every $x\in M$, $x\in N$. Note that
$$N\setminus M = \{x\in N: x\notin M\}$$
Suppose $x\in M \cup N\setminus M$. Then, $x\in M$ or $x\in N\setminus M$. If $x\in M$, then $x\in N$ by $M\subset N$. If $x\in N\setminus M$, then $x\in N$ by definition of $N\setminus M$. So, $M\cup N\setminus M \subset N$.
Next, suppose $x\in N$. Then, either $x\in M$ or $x\notin M$. If $x\in M$, $x\in M\cup N\setminus M$ by definition of the union. If $x\notin M$, then clearly $x\in N\setminus M$, since we assumed $x\in N$, to begin with (refer to the definition of $N\setminus M$ again). As a result, $x\in M\cup N\setminus M$ by definition of the union, once again. So, we have $N\subset M\cup N\setminus M$.
$M\cup N\setminus M \subset N$ and $N\subset M\cup N\setminus M$ together, yield $M\cup N\setminus M = N$. So we have shown $$M\subset N\implies M\cup N\setminus M = N$$ Done!

Lastly, the following image says more than words ever can:
So clearly if $x\in N\setminus M$ then $x\in N$ and if $x\in M$ then $x\in N$ from the hypothesis $M\subseteq N$ so that
$$M\cup N\setminus M\subseteq N$$
and thus we have only to show the other inclusion. So clearly if $x\in N$ then either $x\in M\cap N=M$ or $x\notin M\cap N=M$ (any event is true or false!) and thus $x\in M$ or $x\in N\setminus M$ so that $x\in M\cup N\setminus M$  and thus the statement follows immediately.