Definition:
A maximal element in the partially ordered set (for the inclusion relation) of all filters on a set $X$ is called an ultrafilter on $X$
Theorem:
If $\mathcal{F}$ is a filter on a set $X$, then there exists an ultrafilter $\mathcal{U}$ such that $\mathcal{F} \subseteq \mathcal{U}$
Proof my book provides:
Consider the set $\mathcal{X} := \{\mathcal{G} \mid \mathcal{G}\mathrm{\ filter}, \mathcal{F} \subseteq \mathcal{G}\}$. Then $(\mathcal{X}, \subseteq)$ is a poset and it is inductive ordened. Hence, the existence of the ultrafilter $\mathcal{U}$ follows by Zorn's lemma. $\square$
My questions:
1) I already proved that the set of all filters on $X$ is partially ordered. Does this imply that the poset ($\mathcal{X}, \subseteq)$ is inductive ordened as well? I think not, but my book claims it does.
2) The proof in my book gives a maximal element of the set $\mathcal{X}$, which is a subset of the set of all filters. Hence, is the maximal element found by Zorn's lemma also a maximal element in the set of all filters on $X$? This should be true (but I can't prove it), in order for $\mathcal{U}$ (that is found in the proof) to be an ultrafilter by definition.
I don't know what is the definition of inductive order in that book, but in Wolfram one possibility is a partially ordered set in which every totally ordered subset has an upper bound.
This is the case of $(\mathcal X,\subseteq)$ because if $(\mathcal G_i)_{i \in I}$ is a chain of filters containing $\mathcal F$, then $\mathcal G=\bigcup_{i\in I}\mathcal G_i$ is an upper bound of that chain.
So you are in conditions of applying Zorn's Lemma, and conclude that $(\mathcal X,\subseteq)$ has a maximal element which is, therefore, an ultrafilter.