Is the following subset of a plane connected? (picture)

It's the union of a sequence of interlocked "chains" formed from closed semicircles. The chains can be seen having different shades of gray in the picture.

There's no line in the middle, just the endpoints of arcs which form the semicircles, else it would be trivial.

Connected means it can't be decomposed into 2 nonempty open sets.

I think it's not connected, so I thought about taking one semicircle and including all semicircles which intersect small neighborhoods of the original semicircle's endpoints, iterating this process I can get an open subset which is not the entire thing (if the neighborhoods are chosen sufficiently smaller and smaller). But I would need it also to be closed in order to separate it, but I don't see that it has to be.

This set, which I will denote $S$, is not connected, no matter how you define it (as long as the closed arcs are pairwise disjoint), here is a proof.

Form an equivalence relation $\sim$ on $R^2$ with equivalence classes being the closed arcs you drew and singletons not contained in such arcs.

Then take $X=R^2/\sim$ with the quotient topology, $q: R^2\to X$ is the quotient map. Since every equivalence class is a closed subset of $R^2$, the space $X$ is $T_1$. Then it is easy check that the decomposition of $R^2$ we defined is "upper semicontinuous". This implies that the quotient space is normal (since $R^2$ is normal). See page 8-10 in Daverman's book for definitions and proofs. Incidentally, our space $X$ is homeomorphic to $R^2$, this is a special case of Moore's theorem. We will not need this, however.

Let $T=q(S)\subset X$ denote the projection of $S$. Then $T$ is countable and infinite (this is where I use the fact that all closed arcs are pairwise disjoint). Since $X$ is $T_4$, you can separate any two points $t_1, t_2\in T$ by a continuous function $f: X\to {\mathbb R}$: $f(t_1)\ne f(t_2)$. Since $f(T)$ is countable, we can find a point $r\in {\mathbb R}$ between $f(t_1), f(t_2)$, which is not in $f(T)$. Then $U_1=(fq)^{-1}((-\infty, r))$, $U_2=(fq)^{-1}((r, \infty))$ are disjoint open subsets of $R^2$ separating $q^{-1}(t_1), q^{-1}(t_2)$. The sets $U_1\cap S, U_2\cap S$ are disjoint and their union is the entire $S$. Therefore, we just proved that any two points in $S$ which are not on the same arc, belong to distinct connected components of $S$. In particular, $S$ is disconnected and connected components are the circular arcs.