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: $$\|(x,y)-(a,b)\|< \delta \implies \|B(x,y)-B(a,b)\|< \epsilon.$$


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 $.
