Upper Closure of Element without Element is Filter implies Element is Meet Irreducible

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

Let $x \in S$.

Let
 * $x^\succeq \setminus \left\{ {x}\right\}$ be a filter in $L$.

Then $x$ is meet irreducible.

Proof
Let $a, b \in S$.

Aiming for a contradiction suppose that
 * $x = a \wedge b$ and $x \ne a$ and $x \ne b$

By Meet Precedes Operands:
 * $x \preceq b$ and $x \preceq a$

By definition of upper closure of element:
 * $b, a \in x^\succeq$

By definitions of singleton and difference:
 * $b, a \in x^\succeq \setminus \left\{ {x}\right\}$

By definition of filtered:
 * $\exists z \in x^\succeq \setminus \left\{ {x}\right\}: z \preceq a \land z \preceq b$

By definition of infimum:
 * $z \preceq x$

By definition of upper set:
 * $x \in x^\succeq \setminus \left\{ {x}\right\}$

Thus this contradicts $x \in \left\{ {x}\right\}$ by definition of singleton.