If Y is complete then B(X,Y) is complete I am reading the proof for this theorem and it goes as follows:
To see that $B(X,Y)$ is complete when $Y$ is complete, let $\left\{ T_n \right\}$ be a Cauchy sequence (of operators) in $B(X,Y)$, that is,
$$\|\ T_n - T_m \|\ \to 0 \hspace{.5cm} \text{as} \hspace{.5cm} n,m \to \infty.$$
Then, for each $x \in X$,
$$\|\ T_n x - T_m x \|\ = \|\ (T_n - T_m)x \|\ \leq \|\ T_n - T_m \|\ \|\ x \|\ \to 0 \hspace{.5cm} \text{as} \space\ n,m \to \infty$$
which imples that $\left\{ T_n x \right\} $ is a Cauchy sequence in $Y$. As $Y$ is complete, $\left\{ T_n x \right\}$ converges to some vector, denoted by $Tx$ in $Y$. We have just defined a transformation $T: X \to Y$ by $x \mapsto Tx = \displaystyle\lim_{n\to\infty} T_n x$. Linearity of $T$ is clear. Next, we show that $T \in B(X,Y)$ and $T_n \to T$ with respect to the operator norm. For any $\epsilon > 0$ and any $x \in X$:
\begin{align*}
\|\ (T_n - T)x \|\ &= \|\ (T_n - T_m)x + (T_m - T)x \|\ \\
&\leq \|\ (T_n - T_m)x \|\ + \|\ (T_m - T)x \|\ \\
&\leq \|\ T_n - T_m \|\ \|\ x \|\ + \|\ T_m x - Tx \|\ \\
&< \dfrac{\epsilon}{2} \|\ x \|\ + \dfrac{\epsilon}{2} \|\ x \|\ \hspace{1cm} \exists N \in \mathbb{N} \space\ \text{when} \space\ m,n > N \\
&= \epsilon \|\ x \|\ \\
\end{align*}
Therefore, $\|\ T_n - T \|\ < \epsilon$ whenever $n \geq N$. This shows that $T_n \to T$ in operator norm, and that $\|\ T_n - T \|\ \to 0$ as $n \to \infty$, and so $T$ is continuous. 
My question: Why is $\|\ T_m x - Tx \|\ < \dfrac{\epsilon}{2} \|\ x \|\ $ ? That is the only part of this proof that baffles me. 

For simplicity assume $\|x\| = 1$. From the proof you suggested above two following things hold: 


*

*$\forall x\in X$ with $\|x\|=1$ and $\forall \varepsilon > 0: \exists N_{\varepsilon}$ such that $\forall n,m\geq N_{\varepsilon}$ $\|T_nx-T_mx\|<\varepsilon$

*$\forall x\in X$ with $\|x\|=1$ and $\forall \varepsilon > 0: \exists M_{\varepsilon,x}$ such that $\forall m\geq M_{\varepsilon,x}$ $\|T_mx-Tx\|<\varepsilon$


For $x\in X$ with $\|x\| = 1$ and $\varepsilon >0$ define $m(x) = \text{max}(N_{\varepsilon},M_{\varepsilon,x})$. If $n\geq N_{\varepsilon}$ then from the statements above it follows $$\|T_n x - Tx\| \leq \|T_n x - T_{m(x)}x\|+\|T_{m(x)} x - Tx\| <2\varepsilon.$$ Since this is true for any $x$ with $\|x\|=1$ it follows that $T_n\to T$ in $B(X,Y)$.
