Is every simple continuous curve on a plane a sub-curve of a simple continuous closed curve? Let $P:[0,1]\to \mathbb{R}^2$ be a simple continuous curve on the plane, i.e. $P$ is continuous and injective. Can we always extend it so that it is a sub-curve of a simple closed curve?
It's equivalent to showing that $\mathbb{R}^2\backslash P((0,1))$ is path-connected.
I am only interested in the two dimensional case, but feel free to present a proof with $\mathbb{R}^n,n\geq 2$ replacing $\mathbb{R^2}$.
The fact that $\mathbb{R}^2\backslash P([0,1])$ is path-connected should be useful, see The complement of a simple non closed curve is path connected?.
 A: After surfing over stackexchange I have found an answer.
Let $n\geq 2$. Given that every path-connected Hausdorff space is simple path-connected, see Does path-connected imply simple path-connected?, it suffices to show that $\mathbb{R}^n\backslash P((0,1))$ is path-connected.
Using results in The complement of a simple non closed curve is path connected?, we have that $\mathbb{R}^n \backslash P([0,1])$ is path-connected.
Let $A:=\mathbb{R}^n \backslash P([0,1])$, it suffices to show $A\cup \{P(0),P(1)\}$ is also path-connected.
We need to use the following lemma from When is the closure of a path connected set also path connected?.
Lemma: Let $S\subset X$ be path-connected and $x\in \overline{S}$. Suppose there exists a countable decreasing (i.e. $U^{i+1}\subset U^i$) neighborhood basis $(U_i)_{i=1}^\infty$ in $X$ at $x$ such that for each $i$, whenever $s_i\in S\cap U_i$ then there exists a path in $S\cap U_i$ from $s_i$ to some element of $S\cap U_{i+1}$. Then $S\cup \{x\}$ is path-connected.
Apply the above lemma twice to $A$ then we have the desired result.
