Solving $\lfloor {2x} \rfloor + \lceil {x} \rceil - 4x = 0$ 
I need to find:
$A=\{x \in \mathbb{R}\vert\,\lfloor {2x} \rfloor + \lceil {x} \rceil - 4x = 0\}$

Let $x \in \mathbb{R}$. Then
$$\begin{align*}
\lceil {x} \rceil =4x- \lfloor {2x} \rfloor
&\iff
4x- \lfloor {2x} \rfloor -1       <     x         \leq        4x -\lfloor {2x} \rfloor\\
&\iff
\bigg[ 4x- \lfloor {2x} \rfloor -1      <      x   \bigg]       \land      \bigg[ x       \leq      4x -\lfloor {2x} \rfloor  \bigg]\\
&\iff 
(3x-1      <       \lfloor {2x} \rfloor ) \land (\lfloor {2x} \rfloor       \leq      3x)
\end{align*}
$$
Now define $B=\{x\in \mathbb R\mid(3x-1<\lfloor {2x} \rfloor \}$ and $C=\{x\in\mathbb R\mid\lfloor {2x} \rfloor       \leq      3x\}$. Then $A=B\cap C$.

My confusion:
$-\dfrac{2}{3} \in (B \cap C)$ but $-\dfrac{2}{3} \notin A$
In which step have I gone wrong?


"In which step have I gone wrong?" In the first one: The left hand side implies that $x$ is divisible by $4$ while the right side does not.
Here is how I would approach it.
Notice that $\lfloor 2x \rfloor$ and $\lceil x\rceil$ are always whole numbers so that $\lfloor 2x \rfloor+\lceil x\rceil -4x$ is a whole number if and only if $x=n/4$ for some $n\in\mathbb Z$. Now assume $x=n/4$ is in $A$. Then
$$\lfloor 2x \rfloor+\lceil x\rceil -4x=0$$
which for $n$ means
$$\lfloor n/2 \rfloor+\lceil n/4\rceil =n.$$
Now notice
$$
n/2-1+n/4\le\lfloor n/2 \rfloor+\lceil n/4\rceil \le n/2+ n/4+1
$$
and thus
$$
n/2-1+n/4\le n\le n/2+ n/4+1
$$
which by multiplying with $4$ yields
$$
(3n-1\le 4n\le 3n+1) \iff (-4 \le n \le +4)
$$
Thus the only possible values for $n$ are $\{-4,\dots,4\}$, which are the values $\{-1,-1+1/4,-1+2/4,\dots,1\}$ for $x$. (You still have to check which of these values are in $A$. But after that you are done)
