Proving that the product of two numbers (in $\mathbb{R}$ or $\mathbb{C}$) is a continuous function. This is what is given in the textbook, I will highlight what is confusing me: Product in  field $\mathbb R$ or $\mathbb C$,on $X \times X$ defined as:
$$(x,y)\mapsto xy$$
(Let indicate that map with $B$), we have:
$$\|B(x,y)-B(a,b) \|=\|B(x-a,b)+B(a,y-b)+B(x-a,y-b)\| \\ \leq\|B\|\|x-a\|\|b\|+\|B\|\|a\|\|y-b\|+ \|B\|\|x-a\|\|y-b\|$$

$$\leq\|B\|(2\|(a,b)\|\|(x-a,y-b)\|+\|(x-a,y-b)\|^2)=\|B\|\|(x,y)-(a,b)\|(2\|(a,b)\|+ \|(x,y)-(a,b)\|)$$

Where $\|B\|$ signifies the norm of product and $\|\ x \|=\| \ x \|_{\infty}= \max_{1 \leq i \leq k} \|x^i\|_i$, product product of norms on $X^1 \times X^2$.

We take a $\delta$ such that at the same time $0< \delta <1 $ and $(2\|(a,b)\|+1)\delta < \epsilon $ are true. From there: $$\|(x,y)-(a,b)\|< \delta \implies \|B(x,y)-B(a,b)\|< \epsilon.$$

 A: For the first highlighted step, note that
\begin{align*}
\| x - a \| &\le \max(\|x - a\|, \|y - b\|) = \|(x,y) - (a,b)\| \\
\| y - b \| &\le \max(\|x - a\|, \|y - b\|) = \|(x,y) - (a,b)\| \\
\| a \| &\le \max(\|a\|, \|b\|) = \|(a,b)\| \\
\| b \| &\le \max(\|a\|, \|b\|) = \|(a,b)\|.
\end{align*}
Therefore,
\begin{align*}
\| B \| \| x - a \| \| b \| &\le \| B \| \| (x, y) - (a,b) \| \|(a,b) \| \\
\| B \| \| a \| \| y - b \| &\le \| B \| \|(a,b) \| \| (x, y) - (a,b) \| \\
\| B \| \| x - a \| \| y - b \| &\le \| B \| \| (x,y) - (a,b) \|^2
\end{align*}
So they conclude that, summing the above three inequalities,
$$
\|B(x,y) - B(a,b) \| \le \| B\| \|(x,y) - (a,b) \| \Big[ 2 \|(a,b)\| + \|(x,y) - (a,b) \| \Big]. \tag{1}
$$
Now, fix $\epsilon > 0$. Choose $\delta$ such that $0 < \delta < 1$ and $2 (\|(a,b)\| + 1) \delta < \epsilon$. If you like, you can instead define $\delta$ explicitly:
$$
\delta := \tfrac12 \min \left( 1, \frac{\epsilon}{\|(a,b)\| + 1} \right)
$$
Note, that this is OK since $\delta$ is allowed to depend on the point in question, $(a,b)$. We don't need to show $B$ is uniformly continuous; just continuous at $(a,b)$.
Anyway, take $(x,y)$ such that $\|(x,y) - (a,b)\| \le \delta$. From (1), we then get
\begin{align*}
\|B(x,y) - B(a,b) \|
&< \| B \| \delta \Big[2 \|(a,b)\| + \delta\Big] \\
&< \| B \| \delta \Big[2 \|(a,b)\| + 1\Big] \\
&< \| B \| \epsilon.
\end{align*}
We require one more step--that $\|B\|$ exists and is finite--to then conclude continuity of $B$.
In fact, $\|B\| = 1$.
This is left as an excercise.
Given this, we finally conclude
$$
\|B(x,y) - B(a,b) \|
< \| B \| \epsilon
= \epsilon,
$$
as required.
A: Maybe you can try using sequential continuity, equivalent to continuity in metric (gen. 1st countable) :
Let $x_n \rightarrow x , y_n \rightarrow y$ , we want to show $x_ny_n \rightarrow xy$ :
We have that $(x_n-x)(y_n-y) \rightarrow 0$ as $n \rightarrow \infty$ , since each of the terms on the left can be made small -enough. Now:
$x_ny_n -xy= x_ny_n -x_ny+ x_ny -xy= x_n(y_n-y)+ y(x_n -x)$ Then , as $n \rightarrow  \infty$ we have 
$$\lim_{n \rightarrow \infty} {x_n(y_n-y)+y(x_n-x)} =x(y_n-y)+y(x_n-x)=0,$$ 
and we conclude $ x_ny_n \rightarrow xy $.
