We check each relation property carefully.

1. Recall definitions

For a relation $R$ on $\mathbb{R}$:

• Reflexive: $(x,x)\in R$ for all $x$.
• Symmetric: $(x,y)\in R \Rightarrow (y,x)\in R$.
• Transitive: $(x,y)\in R$ and $(y,z)\in R \Rightarrow (x,z)\in R$.

We need to find the option that is not correct.

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2. Check option A

Relation:

$$(x,y)\in R \iff 0<|x|-|y|\le 1$$

This means

$$|x|>|y|.$$

Symmetry check

If

$$0<|x|-|y|\le 1,$$

then

$$|y|-|x|<0,$$

so $(y,x)\notin R$. Hence it is not symmetric.

Transitivity check

Take

$$|x|=2.0,\quad |y|=1.2,\quad |z|=0.5.$$

Then

$$|x|-|y|=0.8 \le 1,$$

so $(x,y)\in R$. Also,

$$|y|-|z|=0.7 \le 1,$$

so $(y,z)\in R$. But

$$|x|-|z|=1.5>1,$$

so $(x,z)\notin R$. Thus it is not transitive.

So option A is correct.

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3. Check option B

Relation:

$$(x,y)\in R \iff 0<|x-y|\le 1$$

Symmetry check

Since

$$|x-y|=|y-x|,$$

if $0<|x-y|\le 1$, then also $0<|y-x|\le 1$. Hence the relation is symmetric.

Transitivity check

Need to test whether

$$0<|x-y|\le 1,\quad 0<|y-z|\le 1$$

implies

$$0<|x-z|\le 1.$$

This is false in general.

Take

$$x=0,\quad y=1,\quad z=2.$$

Then

$$|x-y|=|0-1|=1,$$

so $(x,y)\in R$, and

$$|y-z|=|1-2|=1,$$

so $(y,z)\in R$. But

$$|x-z|=|0-2|=2>1,$$

so $(x,z)\notin R$. Thus it is not transitive.

So the statement “symmetric and transitive” is incorrect.

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4. Check option C

Relation:

$$(x,y)\in R \iff |x|-|y|\le 1$$

Reflexive check

For any $x$,

$$|x|-|x|=0\le 1,$$

so $(x,x)\in R$ for all $x$. Hence it is reflexive.

Symmetry check

Take $x=3, y=1$. Then

$$|x|-|y|=3-1=2 \not\le 1,$$

so $(x,y)\notin R$. But

$$|y|-|x|=1-3=-2\le 1,$$

so $(y,x)\in R$. Therefore symmetry fails. Hence it is not symmetric.

So option C is correct.

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5. Check option D

Relation:

$$(x,y)\in R \iff |x-y|\le 1$$

Reflexive check

For any $x$,

$$|x-x|=0\le 1,$$

so $(x,x)\in R$. Hence reflexive.

Symmetry check

Since

$$|x-y|=|y-x|,$$

if $(x,y)\in R$, then $(y,x)\in R$. Hence symmetric.

So option D is correct.

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6. Conclusion

Only option B is not correct, because the relation

$$0<|x-y|\le 1$$

is symmetric but not transitive.

Therefore, the required answer is:

$$\boxed{B}$$

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7. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So they agree.