What is semantic closure?
In this context, $F^+ = \{ X \rightarrow Y \mid F \models X \rightarrow Y \}$ is the set of all the dependencies that semantically (logically) follow from $F$ as you wrote.
You also have a syntactic closure, defined as $F^\star = \{ X \rightarrow Y \mid F \vdash X \rightarrow Y \}$ where $F \vdash X \rightarrow Y$ means that $X \rightarrow Y$ can be deduced from $F$ according to an inference system, in this case, Armstrong's axioms.

Armstrong's system is both sound and complete so $F^+ = F^\star$, but the concepts are different. To illustrate the difference, let $\Sigma = \{A \rightarrow B, BC \rightarrow D\}$ and prove the following: 


$AC \rightarrow D \in \Sigma^+$ (semantic proof). Let $r$ an instance such that $r \models A \rightarrow B$ and $r \models BC \rightarrow D$. Assume $t_1, t_2 \in r$ with $t_1[AC] = t_2[AC]$ this implies $t_1[A] = t_2[A]$ which implies $t_1[B] = t_2[B]$ because $r \models A \rightarrow B$. Thus we have $t_1[BC] = t_2[BC]$ and we can conclude $t_1[D] = t_2[D]$ because  $r \models BC \rightarrow D$. Thus $t_1[AC] = t_2[AC]$ implies $t_1[D] = t_2[D]$ for arbitrary tuples $t_1$ and $t_2$ so $r \models AC \rightarrow D$. We have shown an arbitrary instance model of $\Sigma$   is a model of  $AC \rightarrow D$ too, so $\Sigma \models AC \rightarrow D$.
$AC \rightarrow D \in \Sigma^\star$ (axiomatic proof). From  $A \rightarrow B$ we obtain  $AC \rightarrow BC$ by augmentation. Now using $BC \rightarrow D$ we obtain $AC \rightarrow D$ by transitivity.