[proofplan] We prove the theorem by induction on the cardinality of the finite poset. The key finite-poset fact needed in the inductive step is that every nonempty finite poset has a minimal element; we prove this directly to avoid relying on an external result. After choosing a minimal element, we linearly extend the induced subposet on the remaining elements by induction and then place the chosen minimal element first. Minimality ensures that this new total order respects every original comparison involving that element, while the induction hypothesis handles all comparisons among the remaining elements. [/proofplan] [step:Prove that every nonempty finite poset has a minimal element] Let $(Q,\le_Q)$ be a nonempty finite poset. We prove that there exists $q\in Q$ such that no element $r\in Q$ satisfies $r<_Q q$, where $r<_Q q$ means $r\le_Q q$ and $r\ne q$. Choose an element $q_0\in Q$. If $q_0$ is minimal, we are done. If not, there exists $q_1\in Q$ with $q_1<_Q q_0$. If $q_1$ is minimal, we are done. Continuing this process, either it stops at a minimal element or it produces a sequence $(q_i)_{i\ge 0}$ in $Q$ such that \begin{align*} q_{i+1}<_Q q_i \end{align*} for every integer $i\ge 0$. In the second case, the elements $q_i$ are pairwise distinct. Indeed, if $q_j=q_i$ for some $0\le i<j$, then transitivity gives \begin{align*} q_j<_Q q_i, \end{align*} which becomes $q_i<_Q q_i$, contradicting the antisymmetry and irreflexivity of the strict order $<_Q$. Thus the sequence would contain infinitely many distinct elements of the finite set $Q$, impossible. Therefore the process must stop, and $Q$ has a minimal element. [guided] We need a minimal element in the induction step, so we prove the required finite-poset fact here. Let $(Q,\le_Q)$ be a nonempty finite poset. A minimal element means an element $q\in Q$ such that there is no $r\in Q$ with $r<_Q q$, where $r<_Q q$ denotes the conjunction of $r\le_Q q$ and $r\ne q$. Start with any element $q_0\in Q$, which exists because $Q$ is nonempty. If $q_0$ is minimal, the proof is finished. If $q_0$ is not minimal, then by definition there is an element $q_1\in Q$ such that $q_1<_Q q_0$. If $q_1$ is minimal, we are finished. If not, choose $q_2\in Q$ with $q_2<_Q q_1$. Thus, unless we eventually find a minimal element, we construct a sequence $(q_i)_{i\ge 0}$ in $Q$ satisfying \begin{align*} q_{i+1}<_Q q_i \end{align*} for every integer $i\ge 0$. Why is this impossible in a finite poset? The strict inequalities force all the $q_i$ to be distinct. Suppose instead that $q_j=q_i$ for some integers $0\le i<j$. Since \begin{align*} q_j<_Q q_{j-1}<_Q \cdots <_Q q_i, \end{align*} transitivity of the strict order gives \begin{align*} q_j<_Q q_i. \end{align*} Substituting $q_j=q_i$ gives $q_i<_Q q_i$, which contradicts the definition of a strict order induced by a partial order. Hence the sequence would give infinitely many distinct elements of $Q$. This cannot happen because $Q$ is finite. Therefore the descent process must stop, and it stops exactly at a minimal element of $Q$. [/guided] [/step] [step:Induct on the number of elements of the poset] For each integer $n\ge 0$, let $\mathcal P(n)$ be the assertion that every poset with exactly $n$ elements admits a linear extension. We prove $\mathcal P(n)$ for all $n\ge 0$ by induction on $n$. If $n=0$, the empty relation is a total order on the empty set and vacuously extends the given partial order. If $n=1$, the unique relation on the one-element set is a total order and extends the given partial order. Assume now that $n\ge 2$ and that $\mathcal P(k)$ holds for every integer $k<n$. Let $(P,\le)$ be a poset with $|P|=n$. By the previous step, since $P$ is finite and nonempty, there exists a minimal element $m\in P$. [/step] [step:Apply the induction hypothesis after deleting a minimal element] Define \begin{align*} P' := P\setminus\{m\}. \end{align*} Let $\le'$ be the induced partial order on $P'$, meaning that for $x,y\in P'$, \begin{align*} x\le' y \iff x\le y. \end{align*} Then $|P'|=n-1$, so the induction hypothesis gives a total order $\preceq'$ on $P'$ such that, for all $x,y\in P'$, \begin{align*} x\le' y \implies x\preceq' y. \end{align*} [/step] [step:Place the minimal element first and append the induced linear extension] Define a relation $\preceq$ on $P$ as follows. For $a,b\in P$, declare $a\preceq b$ if and only if one of the following holds: $a=b$, or $a=m$, or $a,b\in P'$ and $a\preceq' b$. This relation is a total order on $P$. Comparability follows because $m$ is declared first and any two elements of $P'$ are comparable under the total order $\preceq'$. Antisymmetry follows from antisymmetry of $\preceq'$ on $P'$ together with the fact that no element of $P'$ is $\preceq$-below $m$. Transitivity follows by checking whether one of the elements is $m$: if $m$ appears as the first element of a comparison, it remains first by definition, and if all elements lie in $P'$, transitivity is exactly transitivity of $\preceq'$. [/step] [step:Verify that the constructed total order extends the original partial order] Let $x,y\in P$ satisfy $x\le y$. We prove that $x\preceq y$. If $x=m$, then $x\preceq y$ by the definition of $\preceq$. If $y=m$, then $x\le m$. Since $m$ is minimal, there is no element strictly below $m$, so $x=m$; hence $x\preceq y$ by reflexivity. Finally, if $x,y\in P'$, then $x\le' y$ by definition of the induced order, so $x\preceq' y$ by the induction hypothesis, and therefore $x\preceq y$ by the definition of $\preceq$. Thus $\preceq$ is a total order on $P$ extending $\le$. This proves $\mathcal P(n)$, and induction gives the result for every finite poset. [/step]