= Solution
Use the common nonzero identity $1$. If $\|h\|<1$, <completeness> and <submultiplicativity> give the <Neumann series>
$$
(1-h)^{-1}=\sum_{n=0}^\infty h^n.
$$
For $a\in G(A)$ and $\|a^{-1}\|\|b-a\|<1$, 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 $\|b^{-1}\|$, and
$$
b^{-1}-a^{-1}=b^{-1}(a-b)a^{-1}
$$
proves <continuity> of inversion. Inversion is its own inverse, hence \b[it is a <homeomorphism> of $G(A)$ onto itself].
The <spectrum of an element> is $\sigma_A(x)=\{\lambda:\lambda1-x\notin G(A)\}$, and its complement is the <algebra resolvent set> $\rho_A(x)$. Openness of $G(A)$ makes the <algebra spectrum> <closed>. For $|\lambda|>\|x\|$,
$$
R(\lambda)=(\lambda1-x)^{-1}=\frac1\lambda\sum_{n=0}^\infty(x/\lambda)^n,
$$
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 $(\lambda1-x)R(\lambda)=1$. Hence \b[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 $(\lambda_j1-x)^{-1}\to(\lambda1-x)^{-1}$. Closedness puts this last inverse in $B$, so $\lambda\in\rho_B(x)$. Therefore \b[$\rho_B(x)$ 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.
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