1. Suppose $C$ is a closed polygon with several path-connected components. Is each of these components also closed?
If $C$ is a closed set that is not a polygon, the claim is not true: this can be seen by this example. There is a closed set with two path-components, each of which is not closed. But, if $C$ is a polygon, maybe the claim is true?
2. The above example is related to the following question: suppose $C$ is a path-connected polygon. Is the closure of $C$ path-connected too?
Again, when $C$ is not a polygon, the claim is not true, as shown by the exact same example. But, if $C$ is a polygon, maybe the claim is true?
3. Is it possible to impose on $C$ a condition slightly more general than polygonality, that still guarantees that claims 1 and 2 will be true?
Since you mentioned polygons, I'll assume we're in $\mathbb R^2$ (note that the approach below does not generalize to higher dimensions). Consider some subset $X\subset\mathbb R^2$. For convenience I'll call it a "Jordan subset" if its boundary $\partial X$ is a Jordan curve. Simple polygons are a type of "Jordan subsets". Polygons can be considered as a finite union of simple polygons.
From the way you phrased your question, I suspect you wanted a formal proof of your various statements, so technically I should prove that my above claim is true. For a simple polygon, I'd say it's not too difficult. But for self-intersecting ones, it depends a little bit on what your formal definition of a polygon is.
From the Jordan curve theorem, $\partial X$ splits $\mathbb R^2\setminus X$ into two connected (path-connected actually) components: one is bounded (the interior of the Jordan curve), the other unbounded (the exterior). The interior of $X$ necessarily coincides with one of these components, and $X$ itself is this interior plus some, or all, points of $\partial X$. Either way the closure of $X$ satisfies $\overline X=X\cup\partial X$.
By the Jordan-Schönflies theorem there exists an homeomorphism $\psi:\mathbb R^2\rightarrow\mathbb R^2$ that maps $\partial X$ to the unit circle, its interior (Jordan curve meaning of interior) to the open unit disk, its exterior to the complement of the closed unit disk. So basically a Jordan curve in $\mathbb R^2$ is really just a circle that was continuously deformed. For any $x\in\overline X$ and $y\in X$ we know there exists a path $\gamma$ in the unit disk (or its complement) that connects $\psi(x)$ to $\psi(y)$. Because $\psi^{-1}$ is continuous, $\psi^{-1}(\gamma)$ is a path connecting $x$ to $y$. It follows that $\overline X$ is path-connected.
From 2, each $X_i$ is path-connected. So if $Y^j$ meets $X_i$, it contains the whole $X_i$. It follows that $Y^j$ is a finite union of some of the $X_i$. A finite union of closed sets is closed, so $Y^j$ is closed.
Just a quick note, the Jordan-Schönflies theorem doesn't hold in higher dimension, so you need another approach or another class of sets altogether.