The tickets-in-a-box model of random variables described at https://stats.stackexchange.com/a/54894/919 provides a helpful way to think about this.
Imagine you have a box full of tickets on which are written various numbers in such a way that a blind draw of one ticket acts like observing $X.$ $W$ is a second number found on every ticket: half the tickets display a $1$ and the other half display a $-1$ (that's the definition of a Rademacher variable).
$X$ and $W$ are independent when observing $X$ gives you no clue about what $W$ is and vice versa. Here is the tickets-in-a-box translation of that intuitive description of independence: no matter what the value of $X$ may be, the proportions of tickets with any particular values of $W$ are always the same. In the terminology of conditional probability this says
$$\Pr(W\in\mathcal{A}\mid X\in\mathcal{B}) = \Pr(W\in\mathcal{A})$$
for any events $\mathcal{A}$ and $\mathcal B.$ Equivalently, multiplying both sides of this equation by $\Pr(X),$ we obtain
$$\Pr(X\in\mathcal{A}\text{ and }W\in\mathcal{B}) = \Pr(X\in\mathcal{A})\Pr(W\in\mathcal{B}).$$
That's the mathematical criterion of independence.
Thus, one way to create dependence is to write $W=1$ on some special half of the tickets. In fact, almost any half will do, because the other half--unless you chose these halves very carefully--will exhibit a different distribution of $X$ values.
As a concrete example, let $X$ have a standard Normal distribution and set $W=1$ when $X$ is positive and $W=-1$ otherwise. $X$ tells us exactly what's on $W,$ so $X$ and $W$ cannot be independent.
X
followingN(0,1)
andW
followingw=+1
andW=-1
, each with probability1/2
, could be not independent. $\endgroup$