Consider $\gamma:[0;1]\to\mathbb{R}^2$ is a regular Jordan curve. Let $W=\mathbb{R}^2\setminus \mathrm{Im}(\gamma)$. $C$ is path component of $W$. Proof $C$ is open subset of $\mathbb{R}^2$ and $\partial C\subset \mathrm{Im}(\gamma)$ where $\partial C$ is boundary of $C$ in $\mathbb{R}^2$.
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Jordan curve is injective continuous mapping $\gamma \colon S^1 \rightarrow \mathbb{R}^2$ (see wiki article with jordan curve definition). This means that the curve $Im(\gamma) \subset \mathbb{R}^2$ is some deformation (possibly trivial) of a circle. In other words, it is curve with the same starting and ending point which does not intersect itself otherwise. Implication of the definition is the Jordan theorem (see the same link as the one above) which states that $\mathbb{R}^2$ is divided by $Im(\gamma)$ in two pieces: 1) the interior of $Im(\gamma)$ and 2) the outer region of $Im(\gamma)$ (make a picture if you don't see this).
Both of the pieces 1) and 2) are path connected because $\mathbb{R}^2$ is path connected and $Im(\gamma)$ does not have any crossovers. They are indeed open subsets in the open ball topology on $\mathbb{R}^2$ because every point is contained in some open neighbourhood which is small enough to be contained in the path component (again, this is true for both 1) and 2)).
To be more specific with the proof, consider either of the path components. Definition: A subset is open if with each point of the subset there is an open neighbourhood of the point, contained in the subset. The basic open subsets of $\mathbb{R}^2$ with the open ball topology are the open discs $$ B^2([x_0,y_0],r) = \{ (x-x_0)^2 + (y - y_0)^2 < r^2 | \ r \in \mathbb{R} \} .$$ Now given a point $p$ in the component, measure its smallest distatnce $d$ from the curve $Im(\gamma)$. Now take the open disc to be centered at $p$ and having the radius smaller then $d$, say $r = \frac{d}{2}$. According to the previous definitions, we have found an open neigbourhood of $p$ contained in the component. QED.
The second question concerning the boundary of the path components is also not hard. The boundary is $Im (\gamma)$: for arbitrary point $p \in Im (\gamma)$ the neighbourhood of $p$ has non-empty intersection with the component and its complement in $\mathbb{R}^2$. Since boundary of a given subset is composed of precisely such points (see the first section of topological boundary wiki article) and every point of $Im(\gamma)$ has this property, we are done. QED.