Closed ideal quotient is a Banach algebra

Coset multiplication $(a+I)(b+I):=ab+I$ is well defined: if $a^{\mathrm{\prime }}=a+x$ and $b^{\mathrm{\prime }}=b+y$ with $x,y\in I$, then $a^{\mathrm{\prime }}{b}^{\mathrm{\prime }}=ab+ay+xb+xy$, and $ay,xb,xy\in I$ because $I$ is a two-sided ideal; hence $a^{\mathrm{\prime }}{b}^{\mathrm{\prime }}-ab\in I$ and $a^{\mathrm{\prime }}{b}^{\mathrm{\prime }}+I=ab+I$.

$A\mathrm{/}I$ is a complex vector space with the quotient norm $\mathrm{\parallel }a+I\mathrm{\parallel }={\inf }_{x\in I}\mathrm{\parallel }a-x\mathrm{\parallel }$, and it is complete, hence a Banach space, by [L3] applied to the closed subspace $I\le A$ under Countable Choice.

$A\mathrm{/}I$ is nonzero: were $1+I=0+I$ then $1\in I$, and then $a=a\cdot 1\in I$ for every $a$, so $I=A$, contradicting that $I$ is proper.

Coset multiplication is complex-bilinear: it is the composition of the bilinear product on $A$ with the linear quotient map, so $(\lambda a+\mu a^{\mathrm{\prime }})b+I=\lambda (ab+I)+\mu (a^{\mathrm{\prime }}b+I)$ and similarly in the second variable.

Submultiplicativity. For $a,b\in A$ and $x,y\in I$ one has $ab-(a-x)(b-y)=ay+xb-xy\in I$, so $(a-x)(b-y)\in ab+I$ and hence $\mathrm{\parallel }ab+I\mathrm{\parallel }\le \mathrm{\parallel }(a-x)(b-y)\mathrm{\parallel }\le \mathrm{\parallel }a-x\mathrm{\parallel }\mathrm{\parallel }b-y\mathrm{\parallel }$; given $\epsilon >0$ choose $x,y\in I$ with $\mathrm{\parallel }a-x\mathrm{\parallel }\le \mathrm{\parallel }a+I\mathrm{\parallel }+\epsilon$ and $\mathrm{\parallel }b-y\mathrm{\parallel }\le \mathrm{\parallel }b+I\mathrm{\parallel }+\epsilon$ (the two infima are approximated independently), which gives $\mathrm{\parallel }ab+I\mathrm{\parallel }\le (\mathrm{\parallel }a+I\mathrm{\parallel }+\epsilon )(\mathrm{\parallel }b+I\mathrm{\parallel }+\epsilon )$ and hence $\mathrm{\parallel }(a+I)(b+I)\mathrm{\parallel }\le \mathrm{\parallel }a+I\mathrm{\parallel }\mathrm{\parallel }b+I\mathrm{\parallel }$ after $\epsilon \downarrow 0$.

The quotient unit is normalized. The coset $1+I$ satisfies $(1+I)(a+I)=a+I=(a+I)(1+I)$ by [L1], so it is a two-sided identity, and $\mathrm{\parallel }1+I\mathrm{\parallel }\le \mathrm{\parallel }1-0\mathrm{\parallel }=1$ since $0\in I$; conversely if $\mathrm{\parallel }1+I\mathrm{\parallel }<1$ there is $x\in I$ with $\mathrm{\parallel }1-x\mathrm{\parallel }<1$, so $x=1-(1-x)$ is invertible by [L4], whence $1=x^{-1}x\in I$ and $I=A$ by [step 1.3], a contradiction. Hence $\mathrm{\parallel }1+I\mathrm{\parallel }=1$.

By [step 1.2] $A\mathrm{/}I$ is a Banach space, by [step 2.1] and [step 1.1] its multiplication is an associative complex-bilinear product (associativity descends from $A$ cosetwise), by [step 2.2] the quotient norm is submultiplicative, and by [step 2.3] the coset $1+I$ is an identity of norm one; together with [step 1.3] this says that $A\mathrm{/}I$ is a nonzero unital complex Banach algebra. The quotient map $q(a)=a+I$ is linear, multiplicative, unital and satisfies $\mathrm{\parallel }q\mathrm{\parallel }\le 1$ with $\mathrm{\parallel }q(1)\mathrm{\parallel }=1$, so $\mathrm{\parallel }q\mathrm{\parallel }=1$.