Bounded operators form a Banach algebra, noncommutative in dimension at least two

$\mathcal{B}(X)$ is an associative complex algebra under composition and pointwise linear structure, with unit $1_X$; by [L1] the norm is submultiplicative and $\mathrm{\parallel }{1}_X\mathrm{\parallel }=1$, and by [L2] with $Y=X$ it is complete; hence it is a unital complex Banach algebra.

The space ${\mathbb{C}}^2$ with the maximum norm is a nonzero complex Banach space by [L3], so the argument of [step 1.1] applies to it and $\mathcal{B}({\mathbb{C}}^2)$ is a unital complex Banach algebra; for the explicit operators $U,V$ on that space $\mathrm{\parallel }U(x,y)\mathrm{\parallel }=\mathrm{\parallel }(y,0)\mathrm{\parallel }=\mathrm{\mid }y\mathrm{\mid }\le \mathrm{\parallel }(x,y)\mathrm{\parallel }$ and $\mathrm{\parallel }V(x,y)\mathrm{\parallel }=\mathrm{\mid }x\mathrm{\mid }\le \mathrm{\parallel }(x,y)\mathrm{\parallel }$, so both are bounded, and $UV(x,y)=U(0,x)=(x,0)$ while $VU(x,y)=V(y,0)=(0,y)$, so $UV\mathrm{\ne }VU$ although $UV$ and $VU$ are the two coordinate projections.

Now let $\dim X\ge 2$ and choose linearly independent $u,v\in X$; put $M:=\mathrm{span}\{u,v\}$, a two-dimensional subspace, and let $P$ be a bounded projection with range $M$ by [L4]. The linear maps $S,T:M\to M$ defined by $S(u)=v$, $S(v)=0$ and $T(u)=0$, $T(v)=u$ are bounded by [L5], and they do not commute, since $ST(u)=S(0)=0$ while $TS(u)=T(v)=u\mathrm{\ne }0$. Then $S\circ P$ and $T\circ P$ are bounded operators on $X$ by [L1], and $P\circ S=S$, $P\circ T=T$ because $S$ and $T$ take values in $M$ while $P$ is the identity on $M$; hence $(S\circ P)(T\circ P)=(S\circ T)\circ P$ and $(T\circ P)(S\circ P)=(T\circ S)\circ P$, and these differ because the first sends $u$ to $(S\circ T)(u)=0$ while the second sends $u$ to $(T\circ S)(u)=u\mathrm{\ne }0$. So $\mathcal{B}(X)$ is noncommutative for every $X$ with $\dim X\ge 2$, while [step 2.1] provides the explicit witness on ${\mathbb{C}}^2$; by [step 1.1] the algebra is unital and Banach.