Counter Example of Path Connection

Question

If $A$ is a path connected subset of the topological space $X$ and $A \subset D \subset \bar{A}$ then $D$ is also path connected.

Well I tried to prove that is true but I reach a point that show me that if $D$ is a closed set we can't affirm that is path connected. So I tried searching for a counter example of it but I don't found one.

Answer 1

Consider the set $A=\{(x,\sin(1/x))\mid x\in(0,1)\}$. This set is path connected. Now, $A\subset D=A\cup\{(0,0)\}\subset \overline{A}$ is connected, but it isn't path connected, since there is no curve in it with $(0,0)$ and $(1/2,\sin(2))$ in its range.

Answer 2

S = { (rcos(t),rsin(t)) ; r = 1 - (1/t) t $\geq$ 1 } it a path connected set and $\bar{S}$ = S $\cup$ S$^1$ and if we try to connect by a continuous path the point (0,0) with (1,1) we can't.

I think this is another way of doing this.