The question (in the title) is from Notes on Set Theory by Moschovakis, 2nd edition chapter 6.
First, I am not exactly sure what they mean by "it has a least element." I guessed that the author means that "iff it has a subset $S$ such that $S$ has a least element and that no element of $S$ is comparable to any other element outside of it." But then, the set containing $a,b,c$ and with its nontrivial relations as only $a \leq b$ and $c \leq b$ is inductive, but it doesn't have a least element according to this interpretation... So I am forced to assume that the author means that "iff it has a subset $S$ such that $S$ has a least element." (this is my interpretation for the rest of this post.) But then, any nonempty $P$ has a least element if we take this interpretation, since given $a \in P$, $a$ is a least element of $\{ a \}$ so that we are required to prove that every finite poset is inductive...
Assuming the second interpretation: I tried induction on the number of elements $n$ of $P = \{x_1,...,x_n \}$. For $n=1$ the statement is true trivially. Assume the result for $k <n$.
Let $S$ be a chain in $P$. If $S \neq P$ then by induction $S$ has a supremum. If $S=P$ then $S':=S - \{ x_n \{$ has a supremum, say $x_1$. Since $S=P$ is a chain, either $x_1 \leq x_n$ or $x_n \ x_1$, and the greater of the two is a shpremum.
Is this solution correct?
Edit: For clarification, the author uses the definition that a poset is inductive iff every chain has a least upper bound; in a chain, every element in a chain is comparable.