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. 
 A: First let me answer your explicit question.
The fact that the sequence $T_{n}x$ converges to $Tx$ implies that for every $\delta > 0$, there exists some $N \in \mathbb{N}$ such that $\| T_{n}x - Tx \| < \delta$ for all $n > N$.
Now set $\delta = \epsilon \|x\| /2$.

Now for your implicit question, which seems more important.
There is a subtle point in this proof. When one shows that the sequence $T_{n}$ converges to $T$, one has to show that for each $\epsilon > 0$, there exists an $n \in \mathbb{N}$, such that $\| T_{n} - T \| < \epsilon$, if $n > N$. Notice that there is no mention of any point $x \in X$.
This is where the second index $m$ comes in.
You have proven that I can choose an $N \in \mathbb{N}$ such that $\| T_{n} - T_{m} \| < \epsilon/2$ for all $n,m > N$. Let this $N$ be fixed.
Now let $x \in X$ be arbitrary and let us compute
\begin{equation}
\| (T_{n} - T)x \|  \leqslant \| T_{n} - T_{m} \| \| x \| + \| T_{m} x - Tx \|.
\end{equation}
Note that this inequality is true for any $m \in \mathbb{N}$.
Now let us stop for a second. At this point we can choose and $M \in \mathbb{N}$, such that $\| T_{m}x - Tx \| < \epsilon \| x \| /2$. Suppose without loss of generality that $M > N$. Hence we obtain that
\begin{equation}
\| T_{n} - T_{m} \| \| x \| + \| T_{m} x - Tx \| < \epsilon \| x \|
\end{equation}
for any $n > N$ and any $m > M$.
Putting these things together we see that
\begin{equation}
\|(T_{n} - T) x \| < \epsilon \| x \|.
\end{equation}
Notice that this holds for any $x \in X$, and no a priori choice of $M$ was needed.
A: 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 $B(X,Y)$.
