Consider the following statement.
Let $L$ a finite distributive lattice. Then, for any $x \in L$, we have $x = \sup \{a \in At(L) : a \leq x\}$}.
Let $X = \{a \in At(L) : a \leq x\}$. It is clear, by definition, that $x$ is an upper bound of $X$. I attempted to prove that the statement is true in the following manner.
Assume $x \neq \sup X$. Then there is some $y \in L$ such that $\forall a \in X : a \leq y$ and $y \leq x$. (If $y \not\leq x$ then $a_j \lor a_k$ is undefined for $a_j, a_k \in X$, which means $L$ is not a lattice and implies a contradiction.) Now, if all atoms in $X$ are inferior to a unique element $y$, we may take the subset $D = X \cup \{0, y\}$, where $0$ is the minimum of $L$.
It is clear that $D$ is a sublattice of $L$, because
$\textit{(a.)}~ a_j \lor a_k = y, a_j \land a_k = 0$, for any pair of atoms $a_j, a_k \in X$,
$\textit{(b.)} ~ y \lor a_j = y, y \land a_j = a_j$.
In other words, the supremum and infimum of each pair is included.
It is straightforward to observe that the diamond lattice is a sublattice of $D$---simply observe its Hasse diagram. But this is a contradiction, because $D$ is a sublattice of $L$ and $L$ is distributive by assumption. Then $x = \sup X$. $\blacksquare$
My issue with this proof is indirect. After writing it, I consulted the theory, and the statement is proved in my book only for Boolean algebras$-$this is, for distributive and complemented finite lattices. Furthermore, the proof provided in the book uses complements to arrive at the conclusion. This had me thinking that perhaps my proof is wrong, and there's some reason I am missing why this could be false if the lattice were not complemented.
Is my proof correct? If not, why?