Consider the following statement.
Let $(P, \leq)$ a poset such that $\sup \{a, b\}$ exists for all $a, b \in P$. Then $\sup S$ exists for any $S \subseteq P$.
I set out to show it is true. The poset $P$ is allowed to be infinite. My only idea is to find a recursive formula $f$ for the supremum.
\begin{align*} f(S = \{a_1, a_2, a_3, a_4, \ldots\}) = \sup ~ \Big\{ \sup ~ \{a_1, a_2\}, \sup ~ \{a_3, a_4\}, \ldots \Big\} \end{align*}
It is "clear" that $f(S) = \sup ~ S$, and it is "clear" that it exists because it is made up of pairs of elements. But I'm having trouble passing from the intuition that this is correct to the rigorous proof. One may attempt this by induction. The base case $\{a_1, a_2\}$ is trivial. Then we assume $f(\{a_1, \ldots, a_k\}) = \sup~ \{a_1, \ldots, a_k\}$ holds. Then
\begin{align*} f(\{a_1, \ldots, a_{k+1}\}) = \sup ~ \Big\{ \sup ~ \{a_1, a_2\}, \ldots, \sup ~ \{a_{n-1}, a_n\}, a_k\Big\} \\ &= \end{align*}
and the inductive hypothesis can not be properly applied. Any ideas?
Your statement is false. Let $P=\Bbb N$ with the usual ordering. Then any pair of elements of $P$ has a supremum (just choose the larger of the two numbers) but the set of all positive even integers has no supremum in $P$. In fact, no infinite set has a supremum in $P$.
If any pair of elements has a supremum, you can show that any non-empty finite subset of $P$ has a supremum, but you won't be able to generalize your proof to infinite subsets.