It is a very simple matter to show linear extensions exist in the finite case: one may proceed by induction. Any finite $n$-element poset has a minimal element$x$ (meaning $y \leq x$ implies $y = x$). By induction the restricted partial order on $X \setminus \{x\}$ admits a linear extension, and then one may simply prepend $x$ to that linear order to complete the inductive step.

The rest is a routine application of compactness for propositional theories. Let $(X, \leq)$ be a partially ordered set, and introduce a signature consisting of constants$c_x$, one for each $x \in X$, and a binary relation$L$. Introduce axioms$\neg(c_x = c_y)$ whenever $x \neq y$ in $X$, and $L(c_x, c_y)$ whenever $x \leq y$ in the poset $X$, and axioms stating that $L$ is a linear order. By the previous paragraph, the resulting theory is finitely satisfiable upon interpreting each $c_x$ as $x$. Hence the theory is satisfiable. Taking any model$M$, and interpreting the constants in $M$, and restricting $L$ to them, we obtain a linear extension on $X$.

References

Edward Marczewski, Sur l’extension de l’ordre partiel, Fundamenta Mathematicae, 16: 386–389 (1930) (pdf)