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Hello dear Ali. I think the answer is yes. consider the closed unit interval $I=[0,1]$, and define the set $K$ as follows:$$K=I\times I -(0,1)\times (0)$$

roughly speaking eliminate the interval $(0,1)$ from the bottom of the unit square.

Now we are to define the base of each point of $K$.

If $(x,y)\neq (0,0) , (1,0)$ define the neighborhoods to be as in the usual Euclidean topology.

If $(x,y)=(0,0)$ define the base to be all the sets $[0,\frac{1}{2})\times (0,\epsilon)$, where $\epsilon>0$.

If $(x,y)=(1,0)$ define the base to be all the sets $(\frac{1}{2},1]\times(0,\delta)$, where $\delta>0$.

It is obvious to see that this new space is Hausdorff. But it is not relatively extremely disconnected. To see this consider any neighborhoods of $(0,0)$ and $(1,0)$. it is intuitive to see that the closure of these neighborhoods intersect each other in some point at the edge $y=\frac{1}{2}$. x=\frac{1}{2}$. 1 Hello dear Ali. I think the answer is yes. consider the closed unit interval$I=[0,1]$, and define the set$K$as follows:$$K=I\times I -(0,1)\times (0)$$ roughly speaking eliminate the interval$(0,1)$from the bottom of the unit square. Now we are to define the base of each point of$K$. If$(x,y)\neq (0,0) , (1,0)$define the neighborhoods to be as in the usual Euclidean topology. If$(x,y)=(0,0)$define the base to be all the sets$[0,\frac{1}{2})\times (0,\epsilon)$, where$\epsilon>0$. If$(x,y)=(1,0)$define the base to be all the sets$(\frac{1}{2},1]\times(0,\delta)$, where$\delta>0$. It is obvious to see that this new space is Hausdorff. But it is not relatively extremely disconnected. To see this consider any neighborhoods of$(0,0)$and$(1,0)$. it is intuitive to see that the closure of these neighborhoods intersect each other in the edge$y=\frac{1}{2}\$.