The implication $\sqrt{(x-2)^2+(y-1)^2+(z-1)^2+(w-3)^2}Given a real number $a>0$ find a $b>0$ such that 
$\sqrt{(x-2)^2+(y-1)^2+(z-1)^2+(ω-3)^2}<b\Longrightarrow |xyzω-6|<a$
I tried the procedure followed in another one of my questions, but it doesn't work.

It seems the following. 
The question seems to be a numerical adjustment of the continuity of multiplication.  Assume that $$\sqrt{(x-2)^2+(y-1)^2+(z-1)^2+(w-3)^2}<b.$$ Then $|x-2|<b$, $|y-1|<b$, $|z-1|<b$, and $|w-3|<b$. 
We have $2\cdot 1\cdot 1\cdot 3=6$ and    
$$|xyzw-6|\le$$ $$|xyzw-xyz\cdot 3|+|xyz\cdot 3-xy\cdot 2\cdot 3|+$$ $$|xy\cdot 2\cdot 3 - x\cdot 1\cdot 2\cdot 3|+
|x\cdot 1\cdot 2\cdot 3- 1\cdot 1\cdot 2\cdot 3|=$$ $$
|xyz|\cdot|w-3|+|xy||z-2|\cdot 3+|x||y-1|\cdot 2\cdot 3+
|x- 1|\cdot 1\cdot 2\cdot 3<$$ $$|xyz|\cdot b+|xy|\cdot b\cdot 3+|x|\cdot b \cdot 6+
b \cdot 6.$$
If we additionally assume that $b<1$ then $1<x<3$,  $0<y<2$, $0<z<2$, and $2<w<4$. Hence 
$$|xyz|\cdot b+|xy|\cdot b\cdot 3+|x|\cdot b \cdot 6+b \cdot 6<
3\cdot 2\cdot 2\cdot b+3\cdot 2\cdot b\cdot 3+3\cdot b \cdot 6+b \cdot 6=
54b.$$
Thus it suffices to put $b=\min\{1, a/54\}$.
Update. Such breaking down is a standard trick, used in analysis and based on the triangle inequality. 
The usage of $\min$ is a standard and simple act to show that $b$ should be small. The smallness is typical here,   because the continuity of function in a point is based on its behavior in small neighborhoods of this point. If we reject this approach, we may obtain a formula for $b$, which uses a polynomials or roots of $a$ and which is more complex than the minimum of a constant and a linear function. 
For instance,  we have $|x|<2+b$, $|y|<1+b$, $|z|<1+b$, and $|w|<3+b$. Then 
$$|xyz|\cdot b+|xy|\cdot b\cdot 3+|x|\cdot b \cdot 6+b \cdot 6<$$
$$(2+b)(1+b)^2\cdot b+(2+b)(1+b)\cdot b\cdot 3+(2+b)\cdot b \cdot 6+b \cdot 6=$$ $$
b(b^3+13b^2+23b+17)<b(b^3+15b^2+75b+125) = b(b+5)^3.$$
Then for the given $a$ we have to find $b$ such that  $b(b+5)^3<a$. It suffices to put $b=\frac a{a+6^3}$. Indeed, 
$$\frac a{a+6^3}\left(\frac a{a+6^3}+5\right)^3<\frac a{a+6^3}6^3<a.$$
We can deal with your next inequality as follows. Put $$f=\sqrt{(x-2)^2+(y-1)^2+(z-1)^2}.$$ For given $a>0$ similarly to the above we find positive $b_1, b_2, b_3$ such that 
$f<b_1$ implies $|x^2yz−4|<a/\sqrt{3},$
$f<b_2$ implies $|xy^2z−2|<a/\sqrt{3},$
$f<b_3$ implies $|xyz^2−2|<a/\sqrt{3}.$
Then if $f<\min\{b_1, b_2, b_3\}$ we have $$\sqrt{(x^2yz−4)^2+(x y^2z−2)^2+(xyz^2−2)^2}<\sqrt{3\cdot a^2/3}=a.$$
If you insist on avoiding $\min$, you can put $b=\frac{b_1b_2b_3}{b_1b_2+b_1b_3+b_2b_3}<\min\{b_1, b_2, b_3\}$.
