The first part of the question has little to do with vector spaces:
A subset $A$ of a complete metric space $(X,d)$ is closed if and only if it is complete.
This is a standard exercise (solution in this question).
On (non-metrizable) topological spaces, say that $A$ is sequentially closed if it contains the limits of all its sequences.
Sequentially closed sets need not be closed
(see here for a standard counterexample)
Additionally,
The sequential closure of $A$, i.e. the set of all limit points of sequences in $A$, need not be sequentially closed
(see here).
Assuming that $X$ is additionally a topological vector space is not of any help:
there exist weakly sequentially closed subsets of a ($\infty$-dim) Hilbert space that are not weakly closed.
(see here).
Concerning the second part of the question, the answer is yes: we can conclude that $S$ is closed since $\mathscr P(\Omega)$ is metrizable.
This is a very particular case of the following fact. A topological space $(X,\tau)$ is Polish if there exists a distance $d$ metrizing $\tau$ and additionally such that $(X,d)$ is complete and separable. Endow $\mathscr P(X)$ with the weak topology (induced by the duality with the space $\mathcal C_b(X)$ of continuous bounded functions on $X$). Then
$(X,\tau)$ is Polish if and only if $\mathscr P(X)$ is Polish.
(see e.g. [A, Thm. 8.9.4])
In the case of $\Omega$ compact in $\mathbb R^n$, the weak topology on $\mathscr P(\Omega)$ coincides with the weak* topology since $\mathcal C_b(\Omega)=\mathcal C_c(\Omega)=\mathcal C(\Omega)$. As a consequence, since $\Omega$ is compact (hence complete and separable, thus Polish), $\mathscr{P}(\Omega)$ is Polish (in particular metrizable), and therefore the conclusion follows from the first part of the answer.
Trivia: if Analyst Alice writes "$\mathbb P(\Omega)$" Probabilist Peggy will read "1".
[A] Bogachev, V.I., Measure Theory, vol. II
