I will refer to Example 4.4.7 of Velleman's 2nd edition book 'How to Prove It'.
Find the smallest set of real numbers $X$ such that $5 \in X$ and for all real numbers $x,y$ if $x \in X$ and $x<y$ then $y \in X$.
We know that we are looking for the smallest element of a family of sets
$\mathscr{F} = \{ X \subseteq \mathbb{R}: 5\in X \quad and \quad \forall x\in\mathbb{R}\forall y \in \mathbb{R} [(x\in X \wedge x<y) \rightarrow y \in X] \}$
where the partial order is given by $S= \{ (X,Y) \in \mathscr{P}(\mathbb{R}) \times \mathscr{P}(\mathbb{R}): X \subseteq Y \}$.
It is stated that since $5 \in X$ and $\forall x\in\mathbb{R}\forall y \in \mathbb{R} [(x\in X \wedge x<y) \rightarrow y \in X]$, then we can say that $\forall y \in \mathbb{R} (5<y \rightarrow y \in X)$. This looks like some kind of universal instantiation implementation, but that is as much as comes to my mind when reading this. I would appreciate if you can provide further explanation as to why it is okay to do this step. Also, I would appreciate if you could help me thinking of examples of sets that belong to $\mathscr{F}$. Thank you in advance for your help.
You asked about the step from (a) $5 \in X$ and (b) $\forall x \in \mathbb{R}\forall y \in \mathbb{R}[(x \in X \wedge x < y) \to y \in X]$ to (c) $\forall y \in \mathbb{R}(5 < y \to y \in X)$. Yes, you should start with universal instantiation. Since statement (b) starts with $\forall x \in \mathbb{R}$, you can plug in any real number for $x$. Plugging in 5 for $x$, you get (d) $\forall y \in \mathbb{R}[(5 \in X \wedge 5 < y) \to y \in X]$. Now, statement (a) tells us that $5 \in X$ is true, so for any real number $y$ the statement $(5 \in X \wedge 5 < y)$ that appears in statement (d) will be true if and only if $5 < y$ is true. So you can replace $(5 \in X \wedge 5 < y)$ with the simpler statement $5 < y$, which turns (d) into (c).