Hartogs pseudoconvexity yields a continuous plurisubharmonic exhaustion

Write $b(z)=-\log {\delta }_{\mathrm{\Omega }}(z)$, with $b\equiv 0$ when $\mathrm{\Omega }={\mathbb{C}}^m$ as specified in [L1], and assume $b$ is plurisubharmonic. The function $q(z)={\sum }_{j<m}\mathrm{\mid }{z}_j{\mathrm{\mid }}^2$ is plurisubharmonic by [L3]. For a proper domain, the distance function ${\delta }_{\mathrm{\Omega }}$ is positive and continuous on $\mathrm{\Omega }$ by [L5]: the triangle inequality makes distance to its nonempty complement $1$-Lipschitz. Thus $b$ is continuous in that case, and the whole-space convention is continuous as well. Therefore [L2] makes $u(z):=\max \{b(z),q(z)\}$ a continuous plurisubharmonic function.

For $c<0$, the sublevel $K_c:=\{z\in \mathrm{\Omega }:u(z)\le c\}$ is empty since $q\ge 0$. For $c\ge 0$, it is closed in $\mathrm{\Omega }$ by continuity and lies in $\{q\le c\}$, so it is bounded. If $\mathrm{\Omega }={\mathbb{C}}^m$, it is closed in the whole space and hence compact by [L4]. For a proper domain, every $z\in K_c$ also satisfies ${\delta }_{\mathrm{\Omega }}(z)\ge e^{-c}$. If a sequence from $K_c$ converges in ${\mathbb{C}}^m$ to $a$, the $1$-Lipschitz distance-to-complement function gives distance at least $e^{-c}$ at $a$, so $a\in \mathrm{\Omega }$; continuity of $u$ then puts $a\in K_c$. Thus $K_c$ is closed in the whole space, bounded and contained in $\mathrm{\Omega }$, hence compact by [L4]. All sublevels are compact, so [L1] gives the promised continuous plurisubharmonic exhaustion.