Let D be the open unit disk, and J a Jordan arc (that is a homeomorph of [0, 1]) that lies in D, except J(0) lies on the boundary of D, say J(0)=1. I would like to see that D\J([0, 1]) is a path connected topological space. Please help, if you can. Thanks!

It's certainly the case that $\mathbb{R}^2\setminus J$ is path connected. So any two points in $D\setminus J$ are joined by a path in $\mathbb{R}^2$ missing $J$. If this path isn't in $D$ it hits the boundary of $J$ but then replace part of the path by an arc of the boundary of $D$. Then one can replace this part of the arc by an arc just inside the boundary (there is $\epsilon>0$ such that any closed arc on the boundary not containing $J(0)$ has distance $>\epsilon$ from $J$). 


I know this is a really old question, but I just happened upon it for some reason. I've always liked these kind of plane topology question, so I think I'll add an answer here. If you are looking for a simple topological proof, then I suggest using the following theorem of Janiczewski's: Let z and w be points in the plane, and let $A$ and $B$ be compact, connected sets. If neither $A$ nor $B$ separates $z$ from $w$, and $A\cap B$ is connected, then $A\cup B$ also does not separate $z$ from $w$. Now, to prove that a Jordan arc $J$ does not separate the plane, take two points $z$ and $w$ in $\mathbb{R}^2\setminus J$, and decompose $J$ into small Jordan arcs, each chosen sufficiently small to ensure that none of these separates $z$ from $w$, and apply Janiczewski's theorem repeatedly. To see that the union of your arc with the unit circle does not separate two points inside the unit disk that are not on your arc, just apply Janiczewski's theorem again. 

