 Solution

Use the common nonzero identity $1$. If $\Vert h\Vert <1$, completeness and submultiplicativity give the Neumann seriesFor $a\in G(A)$ and $\Vert a^{-1}\Vert \Vert b-a\Vert <1$, write $b=a(1+a^{-1}(b-a))$ and apply that series. This proves that the group of invertible elements of a Banach algebra is open. Locally the same series bounds $\Vert b^{-1}\Vert$, andproves continuity of inversion. Inversion is its own inverse, hence it is a homeomorphism of $G(A)$ onto itself.

The spectrum of an element is ${\sigma }_A(x)=\{\lambda :\lambda 1-x\notin G(A)\}$, and its complement is the algebra resolvent set ${\rho }_A(x)$. Openness of $G(A)$ makes the algebra spectrum closed. For $|\lambda |>\Vert x\Vert$,so it is bounded and thus compact. For nonemptiness, suppose $R$ existed everywhere. The local Neumann expansion around each point makes $R$ a Banach-space-valued entire function. For every bounded linear functional $\ell$ on $A$, $\ell (R(\lambda ))$ is entire, bounded on compact discs, and tends to zero at infinity by the displayed series. The Liouville theorem makes it identically zero. The Hahn-Banach theorem separates points, so $R(\lambda )=0$, contradicting $(\lambda 1-x)R(\lambda )=1$. Hence the algebra spectrum is nonempty and compact.

If $B$ is a closed unital subalgebra with the same identity, then ${\rho }_B(x)\subseteq {\rho }_A(x)$. It is relatively open by the Neumann series in $B$. If ${\lambda }_j\in {\rho }_B(x)$ tends to $\lambda \in {\rho }_A(x)$, continuity of inversion in $A$ gives $({\lambda }_{j}1-x{)}^{-1}\to (\lambda 1-x{)}^{-1}$. Closedness puts this last inverse in $B$, so $\lambda \in {\rho }_B(x)$. Therefore ${\rho }_B(x)$ is relatively open and closed in ${\rho }_A(x)$. This proves the required resolvent components in a closed unital subalgebra directly, with no spectral-boundary assumption.