Completely Irreducible and Subset Admits Infimum Equals Element implies Element Belongs to Subset

Theorem
Let $L = \left({S, \preceq}\right)$ be an ordered set.

Let $x \in S$ such that
 * $x$ is completely irreducible.

Let $X \subseteq S$ such that
 * $X$ admits an infimum and $x = \inf X$

Then $x \in X$

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

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

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

We will prove that
 * $X \subseteq q^\succeq$

Let $y \in X$.

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

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

By existence of $q$:
 * $q \preceq y$

Thus by definition of upper closure of element:
 * $y \in q^\succeq$

By Infimum of Upper Closure of Element:
 * $\inf\left({q^\succeq}\right) = q$

By Infimum of Subset:
 * $q \preceq x$

Thus this contrsdicts $x \prec q$