Consider $X \ge 0$ on $(\Omega, \mathcal{F}, P)$, suppose $P(X > 0) > 0$. Let $\tilde{\mathcal{F}} \subset \mathcal{F}$ be $\sigma$ subalgebra. Consider $Y = \mathbb{E}(X | \tilde{\mathcal{F}})$. We want to show that $P(Y > 0) > 0$.
My attempt: consider $\tilde{\mathcal{F}}_R = \{R \cap A, A \in \mathcal{\tilde{F}}\}$, where $R$ is the set of positive values of $X$. $X$ is positive a.s. on all elements of $\mathcal{\tilde{F}}_R$. Suppose that set of $A = \{Y \le 0\}$ has positive measure on $\mathcal{\tilde{F}}_R$, then $$0 < \int_A Y dP \le \int_A 0dP = 0$$ Contradiction.
I have one bad point in my proof. What if $\mathcal{\tilde{F}}_R =\{\emptyset\}$. It might be if $\sigma$ subalgebra is small.
You are making things too complicated. $\int_{\Omega} YdP=\int_{\Omega} XdP>0$. This implies that $P(Y>0)>0$.