About Closure under Resolution
Sure. If $C_1\ni x_i$ and $C_2\ni\neg x_i$ are in $F\restriction I$ (in particular, $x_i$ is unset by $I$), pick clauses $D_1,D_2$ in $F$ which restrict to $C_1$ and $C_2$, respectively. Then $x_i\in D_1$ and $\neg x_i\in D_2$, hence their resolvent $(D_1\let\bez\smallsetminus\bez\{x_i\})\cup(D_2\bez\{\neg x_i\})$ is subsumed by some $D\in F$. If $D$ contains a literal made true under $I$, then so does $D_1$ or $D_2$, contradicting their choice. Thus, $D\restriction I$ is in $F\restriction I$, and it subsumes $(C_1\bez\{x_i\})\cup(C_2\cup\{\neg x_i\})$.