I argue the contrapositive, that if $L$ is a finite lattice that is not semimodular, then $L$ has a join irreducible that is not left modular.
Since $L$ is not semimodular, it contains elements $a, b$ such that $a\wedge b\prec a$ but $b\not\prec a\vee b$. Choose any $c$ such that $b < c < a\vee b$. Now choose any $j\leq a$ minimal for the property that $j\not\leq b$; $j$ is join irreducible. Moreover, $j$ is not left modular, since $(j,c)$ is not a modular pair. To verify this last claim, compute that
$b\vee (j\wedge c) = b\vee (j\wedge a\wedge c) = b\vee (j\wedge a\wedge b) = b$,
$(b\vee j)\wedge c = (b\vee (b\wedge a)\vee j)\wedge c = (b\vee a)\wedge c = c$,
and these are different.