Prove that every non-empty finite subset of a lattice has a least upper bound and a greatest lower bound

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Suppose $(P,\leq)$ is a generic lattice. Then for every $\{a,b\}$ in the relation $ \leq $, the set of its upper bounds is not empty and has a least element, and the set of its lower bounds is not empty and has a least element.

Suppose $ S $ is a non-empty subset of the lattice. Two cases arise:

we can denote the upper bounds and lower bounds of a set with $u(S)$ and $ l(S)$ respectively

if $\{a,b\}$ is included in $ S $, then $u(S) $ is included in $ u(\{a,b \})$ since if $x$ is in $u(S) $, then it is true that $ a \leq x$ and $ b \leq x$. But since $ u(\{a,b \})$ has least element, every set included in it also has a least element, therefore there is a least upper bound also in $ u(S)$

The reasoning for the lower bound is analoguous.

I do not understand how to prove it for { a } included in { a,b }. Could you possibly give me some hint on how to proceed with the solution?

Thanks in advance