Assume two event $\mathcal{E}$ and $\mathcal{F}$ such that $\mathcal{F}\subset \mathcal{E}$. Is this always true: $$\mathbb{E}[X|F]\leq \mathbb{E} [X |\mathcal{E}]\frac{\mathbb{P}(\mathcal{E})}{\mathbb{P}(\mathcal{F})}$$?
My answer: Yes, because:
\begin{align} \nonumber \mathbb{E} [X |\mathcal{F}] \nonumber &= \frac{\mathbb{E} \left[X \mathbf{1}_\mathcal{F}\right]}{\mathbb{P}(\mathcal{F})}\\ \nonumber &\leq \frac{\mathbb{E} \left[X\mathbf{1}_\mathcal{E}\right]}{\mathbb{P}(\mathcal{F})}\\ \nonumber &= \frac{\mathbb{E} \left[X\mathbf{1}_\mathcal{E}\right]}{\mathbb{P}(\mathcal{E})}\frac{\mathbb{P}(\mathcal{E})}{\mathbb{P}(\mathcal{F})}\\ \nonumber &= \mathbb{E} [X |\mathcal{E}]\frac{\mathbb{P}(\mathcal{E})}{\mathbb{P}(\mathcal{F})}\\ \end{align}
Is that correct? Is this a property?
This is false.
Let $Y=-X$. Then if
$$\mathbb{E}[X|\mathcal F]\leq \mathbb{E} [X |\mathcal{E}]\frac{\mathbb{P}(\mathcal{E})}{\mathbb{P}(\mathcal{F})}$$
we must have
$$\mathbb{E}[Y|\mathcal F]\geq \mathbb{E} [Y |\mathcal{E}]\frac{\mathbb{P}(\mathcal{E})}{\mathbb{P}(\mathcal{F})}$$
which is a counterexample.
However if $X$ is restricted to be positive then the statement is true, and your proof is valid. (As carmichael561 points out, the invalid step is to assume $\mathbb{E}[X1_\mathcal F]\leq\mathbb{E}[X1_\mathcal E]$. But this step is valid so long as $X$ is always nonnegative.)