1. Find set $A$

Given $A=\{x\in \mathbb{R}:|x+1|<2\}$

Solve the inequality: $|x+1|<2 \iff -2<x+1<2$ $\iff -3<x<1$

So, $A=(-3,1)$

2. Find set $B$

Given $B=\{x\in \mathbb{R}:|x-1|\ge 2\}$

Solve: $|x-1|\ge 2 \iff x-1\ge 2 \text{ or } x-1\le -2$ $\iff x\ge 3 \text{ or } x\le -1$

So, $B=(-\infty,-1]\cup[3,\infty)$

3. Check each option

Option A: $A-B=(-1,1)$

Here $A-B$ means elements in $A$ but not in $B$.

Since $A=(-3,1),\quad B=(-\infty,-1]\cup[3,\infty)$

Within $A$, the part belonging to $B$ is $(-3,1)\cap (-\infty,-1]=(-3,-1]$. Thus remove this from $A$: $A-B=(-3,1)\setminus (-3,-1]=(-1,1)$

So Option A is true.

Option B: $B-A=\mathbb{R}-(-3,1)$

Compute $B-A$ directly.

From $B=(-\infty,-1]\cup[3,\infty), \quad A=(-3,1)$

Remove from $B$ the part common with $A$. Now, $B\cap A=(-3,1)\cap\big(( -\infty,-1]\cup[3,\infty)\big)=(-3,-1]$

Hence, $B-A = \left(( -\infty,-1]\cup[3,\infty)\right)\setminus (-3,-1]$ $= (-\infty,-3]\cup[3,\infty)$

But $\mathbb{R}-(-3,1)=(-\infty,-3]\cup[1,\infty)$

These are not equal, because $[1,3)$ is included in $\mathbb{R}-(-3,1)$ but not in $B-A$.

So Option B is false.

Option C: $A\cap B=(-3,-1]$

Compute: $A\cap B=(-3,1)\cap\big(( -\infty,-1]\cup[3,\infty)\big)=(-3,-1]$

So Option C is true.

Option D: $A\cup B=\mathbb{R}-[1,3)$

Now, $A\cup B=(-3,1)\cup(-\infty,-1]\cup[3,\infty)$

Combining the first two parts: $(-\infty,-1]\cup(-3,1)=(-\infty,1)$

Therefore, $A\cup B=(-\infty,1)\cup[3,\infty)$

And this is exactly $\mathbb{R}-[1,3)$

So Option D is true.

4. Conclusion

The statement which is NOT true is: $\boxed{\text{B}}$