Conversely, let \(A\) be an open connected set. For any \(a\in A,\) the distance \(r_a\) between \(a\) and the complement of \(A\) – which is a closed set – is positive, hence the disk \(D_a = D(a, r_a)\) is a non-empty subset of \(A\) and \(A = \cup_{a \in A} D_a.\) The set \(A\) is one of its dilations, hence it is path-connected. ‌\(\blacksquare\)

For connected sets: let \(\cup_{a \in \cup \mathcal{A}} D_a\) be a dilation of \(\cup \mathcal{A}.\) We have \[ \bigcup_{a \in \cup \mathcal{A}} D_a = \bigcup_{A \in \mathcal{A}} \cup_{a \in A} D_a. \] For any \(A \in \mathcal{A},\) the set \(\cup_{a \in A} D_a\) is a dilation of \(A,\) hence it is path-connected; the inclusion \(A \subset \bigcup_{a \in A} D_a\) provides \[ \cap \mathcal{A} = \bigcap_{A \in \mathcal{A}} A \subset \bigcap_{A \in \mathcal{A}} \cup_{a \in A} D_a, \] hence the intersection of all \(\cup_{a \in A} D_a\) over \(A \in \mathcal{A}\) is not empty. We may therefore apply the result of the theorem for path-connected sets to the collection \(\{\cup_{a \in A} D_a \; | \; A \in \mathcal{A}\}.\) Our arbitrary dilation of \(\cup \mathcal{A}\) is path-connected, hence \(\cup \mathcal{A}\) is connected. ‌\(\blacksquare\)

Assume on the contrary that \(\gamma:[0,1] \to \mathbb{C}\) is a path of \(\overline{A}\) that joins the points \(a_0 = (2/\pi,1)\) and \(a_{\infty} = (0,0);\) it has to go through every point \[ a_n = (x_n, y_n), \; n \in \mathbb{N} \; \mbox{ where } \; \left| \begin{array}{lll} x_n &=& {1}/{((n+1/2) \pi)} \\ y_n &=& \sin 1/x_n = (-1)^n \end{array} \right. \] in this specific order. Indeed, given some \(t_n \in \left[0,1\right[\) such that \(\gamma(t_n) = a_n,\) we have \(\mathrm{Re}(\gamma(t_n)) = x_n.\) As \(\mathrm{Re}(\gamma(1)) = \mathrm{Re}(a_{\infty}) = 0,\) by continuity of \(t\in[0,1] \mapsto \mathrm{Re}(\gamma(t)),\) there is a \(t_{n+1} \in \left]t_n,1 \right[\) such that \(\mathrm{Re}(\gamma(t_{n+1})) = x_{n+1}.\) Since for any \(x>0,\) there is a unique real number \(y\) such that \((x,y) \in A,\) this yields \(\gamma(t_{n+1}) = a_{n+1}.\) Now, since the sequence \(t_n\) is increasing and bounded from above, necessarily \(|t_{n+1} - t_n| \to 0\) when \(n \to +\infty.\) But on the other hand, for any \(n \in \mathbb{N},\) \[ \begin{split} |\gamma(t_{n+1})- \gamma(t_n)| &= |a_{n+1} - a_n| \\ &\geq |y_{n+1} - y_n| \\ &= 2 \end{split} \] Hence the function \(\gamma,\) despite being continuous and defined on the compact set \([0,1]\) cannot be uniformly continuous, which is a contradiction.

The path-connected components of \(A\) are open and path-connected, hence they are also connected. They are also maximal among the connected sets of \(A\mathrm{:}\) a connected component of \(A\) contains a path-connected component of \(A\) if it contains a point of it; if it were to contain more than one path-connected component, it would be the union several disjoint open sets and hence could not be connected. ‌\(\blacksquare\)

Conversely, if \(A\) is not connected, let \(\cup_{a \in A} D_a\) be a dilation of \(A\) which is not path-connected. It has multiple (path-connected) components; let \(B\) be one of them and \(C\) be the union of all the others. Since every component is non-empty and open, \(B\) and \(C\) are both non-empty and open. Then, the function \(f\) defined by \(f(z)=1\) if \(z \in B\) and \(f(z)=0\) if \(z \in C\) is locally constant; however, it is not constant. ‌\(\blacksquare\)