Let $\mathcal{D} = \{Q \in L^2(\Omega, \mathcal{F}, P) \ |\ \forall \omega \in \Omega \ \ Q(\omega) \geq 0, E[Q] = 1 \}$. I want to show that the set $$\left\{ X \in L^2(\Omega, \mathcal{F}, P) \ |\ \sup_{Q \in D} \{E[QX]\} \leq 0 \right\} $$ is closed.
I have attempted the following: Let $\left( X_k\right)_{k\in \mathbb{N}} \subseteq L^2$ be a sequence of random variables and $X \in L^2$ such that $\|X_k - X\|_2 \rightarrow 0$ and $\sup_{Q \in D} \{E[QX]\} \leq 0$ for each $k$. Then, it follows that \begin{align} 0 & \leq \|X_k-X\|_1 \\ & = E[|X_k-X|]\\ & \leq \|X_k-X \|_2 \rightarrow 0, \end{align} so $$ E[|X_k-X] \rightarrow 0.$$
Appealing to Jensen's inequality we find that \begin{align} |E[X_k] - E[X]| \leq E[|X_k-X|], \end{align} so $$E[X_k] \rightarrow E[X].$$
This only shows that the set in question is closed for the special case where $\mathcal{D} = \{1\}$, i.e, when we only consider the probability measure $P$.
Is there any way forward in this approach? Am I missing some assumptions?
Suppose $EQX \leq 0$ for all $Q \in \mathcal D$. Let $P(E)>0$ and take $Q=\frac 1 {P(E)} I_E$. Then $Q \in \mathcal D$ so $EQX\leq 0$. Hence $\int_EXdP\leq 0$ whenever $P(E)>0$. But this is possible only when $X\leq 0$ a .s. Hence your set contains exactly all $X \in L^{2}$ such that $X\leq 0$ a .s.. It is trivial to see that that this set is closed . [Recall that convergence in $L^{2}$ implies a.s. convergence of a subsequence].