Does $E(Y)=0$ hold in this setting? Given two continuous random variables $X$ and $Y$ with the same support, we have $E(X)=0$ and $Y\leq|X|$. Does this setting imply $E(Y)=0$?
My proof: $Y\leq|X|\Longrightarrow -X\leq Y\leq X \Longrightarrow 0=-E(X)\leq E(Y)\leq E(X) = 0 \Longrightarrow E(Y)=0$. Does this make sense?
 A: $Y \leq |X|$ does not imply that $-X \leq Y \leq X$.
For a counter-example think of the case $Y=X \sin X$ if $X >0$ and $Y=-X$ if $X \leq 0$ with $X \sim N(0,1)$.
A: Counterexample:
$$\begin{align}
X&\sim\text{Normal}(0,1)\\[1ex]
Y&=X\,e^{-X}\\[1ex]
\mathbb E[Y]&=-\sqrt{e}\end{align}
$$
A: The other answers provide good counterexamples saying why your statement cannot be true. I just wanted to go into more detail pointing out the flaw in your logic.
While it is true that $|Y|\le X\implies -X\le Y\le X$, it is not the case that $Y\le |X|$ implies $-X\le Y\le X$. I think this summarizes the mistake you made.
Instead, $Y\le |X|$ implies
$$
Y\le 0\quad \text{or}\quad 0\le Y\le X \quad \text{or}\quad 0\le Y\le -X.
$$
A: If $Y$ and $X$ share the same support but are not necessarily identically distributed, there is no reason why you should have $\operatorname{E}[Y] = 0$.  Moreover, it is not clear what you mean by $Y \le |X|$ because $X$ and $Y$ are random variables.
If what you mean is $$\Pr[Y \le |X|] = 1,$$ then it is still not true:  for example, let $U \sim \operatorname{Uniform}(0,1)$, and then define $$X = U - 1/2, \\Y \mid X = (X + 1/2)B - 1/2,$$ where $B \sim \operatorname{Beta}(2,3)$.  Then $X$ obviously has zero mean, and the unconditional distribution of $Y$ has support on $(-1/2,1/2)$, the same as $X$, since $Y$ attains its maximum when $U = B = 1$ and its minimum when $B = 0$.  Since $Y$ depends on $X$, $$\Pr[Y \le |X|] = \Pr[UB  \le |U - 1/2| + 1/2],$$ and the triangle inequality gives $$|U-1/2| + |1/2| \ge |U - 1/2 + 1/2| = |U| = U \ge UB,$$ hence this probability equals $1$.
But $$\operatorname{E}[Y] = \operatorname{E}[\operatorname{E}[(X + 1/2)B - 1/2 \mid X]] = \operatorname{E}[(X + 1/2)\operatorname{E}[B] - 1/2] = \frac{2}{5}\operatorname{E}[X] - \frac{3}{10} = -\frac{3}{10} \ne 0.$$
