Connected closed subset in a plane Could anyone give a hint on the proof of the following fact?

Let $X$ be a closed connected subset in a a 2-sphere. Then every connected component of the complement of $X$ is simply connected.

It seems to use Jordan curve theorem. If one component is not simply connected, then it contains an immersed loop which is not contractible. But could this immersed loop be improved to be Jordan curve? Any hint? Is it possible to use a polygon to approximate to approach the immersed curve?  
Also is the following true?

A closed subset is connected in a 2-sphere if and only if each component of its complement is simply connected.

 A: Proposition. Let $X$ be a closed subset of the sphere $\mathbb S^2$. Then $X$ is connected if and only if every connected component of $\mathbb S^2\setminus X$ is simply connected.
Necessary condition. Suppose some connected component $U$ of $\mathbb S^2\setminus X$ is not simply connected, that is, there exists a loop $\sigma:[0,1]\to U$ which is not contractible in $U$.
If $\sigma$ was an embedded $\mathbb S^1$ (a Jordan loop), then its complement in $\mathbb S^2$ would consists of two connected components $V,W$. Since $X$ is connected, only one could meet $X$, and the other would be contained in $U$, making $\sigma$ contractible in $U$. And this is the idea, although not that easy, because the loop, being just continuous may be very weird:

Denote $\varSigma$ the trace of $\sigma$ and pick a point $p\in\mathbb S^2\setminus\varSigma$ so that $\varSigma\subset U\setminus\{p\}\subset\mathbb S^2\setminus\{p\}\equiv\mathbb R^2$. The rough argument is to replace the loop by a polygonal joining suitable points of $\varSigma$, which we can do locally by interpolation in small discs $D_i$ covering $\varSigma$. But again, as the picture above tries to show, this may be elusive:
in the case depicted the disc $D_r$ must be repeated as $D_2$ and $D_2$ renamed as $D_3$, and things get worse with other discs.
To do this properly, cover $\varSigma$ with discs $D_i\subset U$. Their inverse images $\sigma^{-1}(D_i)$ form an open covering on $[0,1]$ and it has a Lebesgue number $\rho>0$ such that for any partition $0=t_0<t_1<\cdots<t_r=1$ with $\rho\!>\!1/r\ $ each image $\sigma([t_{k-1},t_k])$ lies in a disc $D_{i_k}\subset U$.
Next the segment $J_k\subset D_{i_k}$ joining $\sigma(t_{k-1})$ to $\sigma(t_k)$ is parametrized by the interval $[t_{k-1},t_k]$ to get a finite polygonal loop $\gamma:[0,1]\to \bigcup D_i\subset U$. This $\gamma$ is homotopic to $\sigma$ by $H_s=(1-s)\sigma+s\gamma$; notice here that linear interpolation remains inside the $D_{i_k}$'s. This guarantees $\gamma$ is not contractible in $U$, as $\sigma$ was not.
Now, $\gamma$ being a finite polygonal loop, it is indeed a finite union of polygonal Jordan loops, and some of them $L$ must be not contractible, hence we have in the end a polygonal Jordan loop $L\subset U$ and for this so simple loop the argument sketched at the beginning does work to get a contradiction.
Notice also that we are using the Jordan-Brouwer theorem, but the cheap case of a polygonal Jordan loop!
Sufficient condition. Suppose $X\subset\mathbb S^2$ is closed but not connected, say $X=Y\cup Z$ were $Y,Z$ are disjoint closed subsets of the sphere. We must construct a non-contractible loop $\sigma:[0,1]\to\mathbb S^2\setminus X$.
Here I will use some Differential Topology.
First we take a smooth function $f:\mathbb S^2\to\mathbb R$ which is $\equiv-1$ on $Y$ and $\equiv1$ on $Z$ (Uryshon's smooth function). Then we approximate $f$ by $g:\mathbb S^2\to\mathbb R$ which is $<0$ on $Y$ and $>0$ on $Z$ and is transversal to $\{0\}\subset\mathbb R$. Consequently, $C=g^{-1}(0)$ is a compact smooth curve, hence a finite union of smooth Jordan curves, say $C=C_1\cup\cdots\cup C_s$; by construction $C\cap X=\varnothing$. Each $C_\ell$ disconnects $\mathbb S^2$ (truly Jordan-Brouwer), hence $C_\ell$ has a regular smooth equation $g_\ell:\mathbb S^1\to\mathbb R$, which is a smooth function transversal to $\{0\}$ such that $C_\ell=\{g_\ell=0\}$. Of course $C_\ell$ has such local equations, but there is a problem with the choice of sign. Now as the curve disconnects one chooses the sign to be constant in a fixed component of the complement, and then the local equations can be glued well by means of a partition of unity. (This is too sketchy, but is the essential thing. Thinking of a non-disconnecting circle in a torus helps to see what's going on.)
An important fact is that any smooth function $h:\mathbb S^1\to\mathbb R$ vanishing on $C_\ell$ is a multiple of the equation $g_\ell$, that is, $h=h_\ell\cdot g_\ell$. Indeed, (i) locally at $C_\ell$ this reduces to the case $g_\ell=y$ in the plane, an easy Calculus exercise, and (ii) off $C_\ell$ one can divide by $g_\ell$ because it has no zero.
It follows that there is a factorization $g=ug_1\cdots g_s$, where $u$ is a smooth function nowhere zero. Thus $u$ has constant sign on $\mathbb S^2$, and since $g$ changes sign form $Y$ to $Z$, some factor $g_\ell$ must do the same. For instance, $g_1$ is $<0$ at some point $p\in Y$ and $>0$ at some point $q\in Z$. Consider the connected component $U$ of $\mathbb S^2\setminus X$ that contains $C_1$ and the inclusions
$$
C_1\subset U\subset\mathbb S^2\setminus\{p,q\}=M. 
$$
If $C_1$ was contractible in $U$ it would be in $M$. But dropping $p$ gives the plane and there the Jordan loop $C_1$ with $q$ in its interior (as Jordan loop). Thus $M$ is a plane minus one point and $C_1$ a loop around that point, hence $C_1$ cannot be contractible in $M$. We are done.
