If (Y) is Banach then (\mathcal B(X,Y)) is Banach

Fix $x\in X$. Since $(T_n)$ is Cauchy in operator norm, [L2] gives $\mathrm{\parallel }{T}_{n}x-T_{m}x\mathrm{\parallel }\le \mathrm{\parallel }{T}_n-T_m\mathrm{\parallel }\mathrm{\parallel }x\mathrm{\parallel }$, so $(T_{n}x)$ is a Cauchy sequence in $Y$. Because $Y$ is Banach, there is $Tx\in Y$ with $T_{n}x\to Tx$.

Step 1.1 defines a map $T:X\to Y$. If $x,y\in X$, then $T_n(x+y)=T_{n}x+T_{n}y$ for every $n$, so passing to the limit gives $T(x+y)=Tx+Ty$. The same argument with $\lambda x$ gives $T(\lambda x)=\lambda Tx$. Thus $T$ is linear.

Choose $N$ such that $\mathrm{\parallel }{T}_n-T_m\mathrm{\parallel }\le 1$ for all $m,n\ge N$. Fix $n\ge N$ and $x\in X$. For every $m\ge N$, $\mathrm{\parallel }{T}_{m}x\mathrm{\parallel }\le \mathrm{\parallel }{T}_{N}x\mathrm{\parallel }+\mathrm{\parallel }(T_m-T_N)x\mathrm{\parallel }\le (\mathrm{\parallel }{T}_N\mathrm{\parallel }+1)\mathrm{\parallel }x\mathrm{\parallel }$ by [L2]. Letting $m\to \mathrm{\infty }$ in step 1.1 gives $\mathrm{\parallel }Tx\mathrm{\parallel }\le (\mathrm{\parallel }{T}_N\mathrm{\parallel }+1)\mathrm{\parallel }x\mathrm{\parallel }$, so $T$ is bounded and hence $T\in \mathcal{B}(X,Y)$.

Let $\epsilon >0$. Since $(T_n)$ is operator-norm Cauchy, choose $N$ so that $\mathrm{\parallel }{T}_n-T_m\mathrm{\parallel }<\epsilon$ for all $m,n\ge N$. Fix $n\ge N$ and $x$ with $\mathrm{\parallel }x\mathrm{\parallel }\le 1$. Step 1.1 gives $T_{m}x\to Tx$, so $\mathrm{\parallel }(T_n-T)x\mathrm{\parallel }={\lim }_m\mathrm{\parallel }(T_n-T_m)x\mathrm{\parallel }\le \epsilon$ by [L2]. Taking the supremum over the unit ball yields $\mathrm{\parallel }{T}_n-T\mathrm{\parallel }\le \epsilon$.

Step 2.3 shows $T_n\to T$ in operator norm, with $T\in \mathcal{B}(X,Y)$ by step 2.2. Therefore every operator-norm Cauchy sequence converges in $\mathcal{B}(X,Y)$, so $\mathcal{B}(X,Y)$ is Banach by [L1].