Supremum inequality over a seperable space

Question

I need some help by understanding the following:

Suppose that $\mathcal{F} = \{f: \mathbb{R}^d \to \mathbb{R} \; measurable\}$ is seperable and $P$ is some probability measure. Then it holds that: $\sup\limits_{f \in \mathcal{F}} P(f) \leq \sup\limits_{S \subset \mathcal{F} countable} P(\sup\limits_{f \in S} f). $

My questions:

1. Does the following equality hold? $\sup\limits_{f \in \mathcal{F}} f = \sup\limits_{S \subset \mathcal{F} countable} \sup\limits_{f \in S} f.$ If so, why is the above inequality not an equality? If not, why isn't it an equality?
2. Why can I put the supremum in front of the measure $P$? Can I do this every time or do I need some conditions on $P$ to do that?
3. I think the inequality has something to do with the fact, that (since $\mathcal{F}$ is seperable) there exists some countable and dense subset $S \subset \mathcal{F}$ such that $\bar{S} = \mathcal{F}$ (here $\bar{S}$ denotes the closure of $S$). Is that true?