Completely Irreducible Element equals Infimum of Subset implies Element Belongs to Subset

Theorem
Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a complete lattice.

Let $X \subseteq S$, $p \in S$ such that
 * $p$ is completely irreducible and $p = \inf X$

Then $p \in X$

Proof
Aiming for a contradiction suppose that
 * $p \notin X$

By Completely Irreducible Element iff Exists Element that Strictly Succeeds First Element:
 * $\exists q \in S: p \prec q \land \left({\forall s \in S: p \prec s \implies q \preceq s}\right) \land p^\succeq = \left\{ {p}\right\} \cup q^\succeq$

where $p^\succeq$ denotes the upper closure of $p$.

We will prove that
 * $q$ is lower bound for $X$

Let $x \in X$.

By definitions of infimum and lower bound
 * $p \preceq x$

By assumption:
 * $x \ne p$

By definition of strictly precede:
 * $p \prec x$

Thus by step condition:
 * $q \preceq x$

By definition of strictly precede:
 * $p \preceq q$

We will prove that
 * $\forall b \in S: b$ is lower bound for $X \implies b \preceq q$

Let $b \in S$ such that
 * $b$ is lower bound for $X$.

By definition of infimum:
 * $b \preceq p$

Thus by definition of transitivity:
 * $b \preceq q$

By definition of infimum:
 * $q = \inf X = p$

Thus this contradicts $p \ne q$ by definition of strictly precede.