Hartogs pseudoconvexity implies Levi pseudoconvexity for $C^2$ domains

Suppose toward a contradiction that $\mathrm{\Omega }$ is not Levi pseudoconvex at a boundary point $p$. Choose a defining function $\rho$ and a complex tangent vector $v$ with ${\mathcal{L}}_{\rho }(p;v)<0$. Translate $p$ to $0$, make a complex-linear change of coordinates sending $v$ to the first coordinate direction and the real normal into the last coordinate, and normalize the last coordinate by subtracting the holomorphic pure-quadratic part of the Taylor expansion. After multiplying $\rho$ by a positive constant, its restriction to the resulting $(z_1,z_m)$-plane has the form $\rho (z_1,0,\dots ,0,z_m)=\mathrm{Re}{z}_m-c\mathrm{\mid }{z}_1{\mathrm{\mid }}^2+o(\mathrm{\mid }{z}_1{\mathrm{\mid }}^2+\mathrm{\mid }{z}_m{\mathrm{\mid }}^2)$ for some $c>0$; subtracting that pure-quadratic part does not change the tangential Levi coefficient. Choose $\lambda >0$ and then $s_0>0$ small enough that the remainder is dominated by the two displayed negative terms. The affine analytic discs ${\mathrm{\Phi }}_s(\zeta )=(\lambda \zeta ,0,\dots ,0,-s)(\mathrm{\mid }\zeta \mathrm{\mid }\le 1)$ then lie in $\mathrm{\Omega }$ for $0<s\le s_0$, their centres ${\mathrm{\Phi }}_s(0)$ converge to $p$ as $s\downarrow 0$, and $K:={\bigcup }_{0\le s\le s_0}{\mathrm{\Phi }}_s(\mathrm{\partial }\mathbb{D})$ is compactly contained in $\mathrm{\Omega }$: on the boundary circles the term $-c{\lambda }^2$ stays uniformly negative even at $s=0$.

By [L1], choose a continuous plurisubharmonic exhaustion $u$ of $\mathrm{\Omega }$, and let $M:={\max }_{K}u$. For $0<s\le s_0$, the function $u\circ {\mathrm{\Phi }}_s$ is subharmonic on $\mathbb{D}$ by [L2] and continuous on its closure. Its boundary values are at most $M$, so [L4] gives $u({\mathrm{\Phi }}_s(0))\le M$. Thus all the centres lie in the compact sublevel set $\{u\le M\}⋐\mathrm{\Omega }$. But those centres converge to the boundary point $p$, contradicting compact containment. Therefore no negative tangential Levi direction exists, and [L3] makes $\mathrm{\Omega }$ Levi pseudoconvex.