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I like my (previous) answer in its simple form. Thus I will expand it here (it should not be a waste, I hope), and first of all I will fulfill Todd's request, see Thm 1 below: we want to show that a closed subset of $\ \mathbb R^2,\ $ which separates $\ \mathbb R^2,\ $ is not totally disconnected. The proof is smoother in the compact case, but in general it needs just a minor extra consideration.

A direct dimension-flavor approach is possible. However I feel like applying the elegant Borsuk's theorem based on his separation criterion. Eilenberg and Steenrod cover this material beautifully in their Chapter 11 (it's like a perfectly elementary appendix) to their "Foundations of Algebraic Topology".

BORSUK's THEOREM   Let $\ X\ $ be a closed subset of $\ S^n.\ $ Then $\ S^n\setminus X\ $ is disconnected $\ \Leftrightarrow\ $ there exists a continuous map $\ f:X\rightarrow S^{n-1}\ $ which is not homotopic to a constant.

On this occasion I also have selected this direction of presentation because Bill Thurston in a respective post wrote about the Alexander duality in a way which harmonized with the above. I read that Bill's post only after Kristal's answer (which didn't really answer) since it referenced to the said Bill's post; the very first comment of this whole thread, just under the QUESTION of the threat, written by me before Kristal's answer, read:

    It is not possible. Where does this problem come from?

Now, in the compact context the being totally disconnected is equivalent to being $0$-dimensional. This means for the totally disconnected (i.e. $0$-dimensional) closed set $\ X\subseteq S^n\ $ every continuous map $\ f:X\rightarrow S^{n-1}\ $ is homotopically trivial for $\ n\ge 2.\ $ Thus for a compact $\ X\subseteq\mathbb R^2\ $ which disconnects $\ R^2\ $ the requested theorem already holds, i.e. $\ X\ $ is not totally disconnected (add $\infty$ to the unbounded component of $\ \mathbb R^2\setminus X).\ $ Thus we have to include in the proof only the (nuisance :-) non-compact case (while the theorem is formulated for both):

THEOREM 1   Let $\ X\ $ be a closed subset of $\ \mathbb R^2\ $ such that $\ \mathbb R^2\setminus X\ $ is not connected. Then $\ X\ $ is not totally disconnected.

PROOF The case of compact $\ X\ $ was covered above. Now let $\ X\ $ be closed and not compact. Then $\ Y := X\cup\{\infty\}\ $ is compact. Of course $\ Y\ $ disconnects $\ \{\infty\}\cup \mathbb R^2 = S^2.\ $ Thus by the Borsuk's theorem and the remarks which followed, $\ \dim(Y)\ge 1.\ $ However $\ X\ $ is $\sigma$-compact, $\ X=\bigcup_{n=1\ 2\ \ldots}X_n\ $ where each $\ X_n\ $ is compact. Thus $\ \dim(\{\infty\}\cup\bigcup_{n=1\ 2\ \ldots}X_n) = \dim Y \ge 1,\ $ and $\ \dim(\{\infty\}) = 0.\ $ Thus $\ \exists_n \dim(X_n)\ge 1.\ $ Such $\ X_n\ $ is not totally disconnected, hence neither is $\ X$.   END of Proof

Todd's request's fulfilled; now back to the generalization, while George Lowther has already provided a still more general result (nice!)

The proof in my first answer used only the assumption that set $\ A\ $ was disconnected, and not more--it didn't mention that $\ A\ $ was totally disconnected. Thus (as promised in my first answer) I really proved:

THEOREM 2   If $\ \mathbb R^2 = A\cup B\ $, and $\ A\ $ is not connected, then $\ B\ $ is not totally disconnected.

In other words:

THEOREM 2'   If $\ \mathbb R^2 = A\cup B\ $, and $\ B\ $ is totally disconnected then $\ A\ $ is connected.

Actually, all this holds for $\ \mathbb R^n\ $ for every $\ n\ge\ 2 (not just for $\ n=2). This follows from the Borsuk's theorem for each $\ n\ge 2\ $ just as it does from its special case of $\ n=2$.

I like my (previous) answer in its simple form. Thus I will expand it here (it should not be a waste, I hope), and first of all I will fulfill Todd's request, see Thm 1 below: we want to show that a closed subset of $\ \mathbb R^2,\ $ which separates $\ \mathbb R^2,\ $ is not totally disconnected. The proof is smoother in the compact case, but in general it needs just a minor extra consideration.

A direct dimension-flavor approach is possible. However I feel like applying the elegant Borsuk's theorem based on his separation criterion. Eilenberg and Steenrod cover this material beautifully in their Chapter 11 (it's like a perfectly elementary appendix) to their "Foundations of Algebraic Topology".

BORSUK's THEOREM   Let $\ X\ $ be a closed subset of $\ S^n.\ $ Then $\ S^n\setminus X\ $ is disconnected $\ \Leftrightarrow\ $ there exists a continuous map $\ f:X\rightarrow S^{n-1}\ $ which is not homotopic to a constant.

On this occasion I also have selected this direction of presentation because Bill Thurston in a respective post wrote about the Alexander duality in a way which harmonized with the above. I read that Bill's post only after Kristal's answer (which didn't really answer) since it referenced to the said Bill's post; the very first comment of this whole thread, just under the QUESTION of the threat, written by me before Kristal's answer, read:

    It is not possible. Where does this problem come from?

Now, in the compact context the being totally disconnected is equivalent to being $0$-dimensional. This means for the totally disconnected (i.e. $0$-dimensional) closed set $\ X\subseteq S^n\ $ every continuous map $\ f:X\rightarrow S^{n-1}\ $ is homotopically trivial for $\ n\ge 2.\ $ Thus for a compact $\ X\subseteq\mathbb R^2\ $ which disconnects $\ R^2\ $ the requested theorem already holds, i.e. $\ X\ $ is not totally disconnected (add $\infty$ to the unbounded component of $\ \mathbb R^2\setminus X).\ $ Thus we have to include in the proof only the (nuisance :-) non-compact case (while the theorem is formulated for both):

THEOREM 1   Let $\ X\ $ be a closed subset of $\ \mathbb R^2\ $ such that $\ \mathbb R^2\setminus X\ $ is not connected. Then $\ X\ $ is not totally disconnected.

PROOF The case of compact $\ X\ $ was covered above. Now let $\ X\ $ be closed and not compact. Then $\ Y := X\cup\{\infty\}\ $ is compact. Of course $\ Y\ $ disconnects $\ \{\infty\}\cup \mathbb R^2 = S^2.\ $ Thus by the Borsuk's theorem and the remarks which followed, $\ \dim(Y)\ge 1.\ $ However $\ X\ $ is $\sigma$-compact, $\ X=\bigcup_{n=1\ 2\ \ldots}X_n\ $ where each $\ X_n\ $ is compact. Thus $\ \dim(\{\infty\}\cup\bigcup_{n=1\ 2\ \ldots}X_n) = \dim Y \ge 1,\ $ and $\ \dim(\{\infty\}) = 0.\ $ Thus $\ \exists_n \dim(X_n)\ge 1.\ $ Such $\ X_n\ $ is not totally disconnected, hence neither is $\ X$.   END of Proof

Todd's request's fulfilled; now back to the generalization, while George Lowther has already provided a still more general result (nice!)

The proof in my first answer used only the assumption that set $\ A\ $ was disconnected, and not more--it didn't mention that $\ A\ $ was totally disconnected. Thus (as promised in my first answer) I really proved:

THEOREM 2   If $\ \mathbb R^2 = A\cup B\ $, and $\ A\ $ is not connected, then $\ B\ $ is not totally disconnected.

In other words:

THEOREM 2'   If $\ \mathbb R^2 = A\cup B\ $, and $\ B\ $ is totally disconnected then $\ A\ $ is connected.

Actually, all this holds for $\ \mathbb R^n\ $ for every $\ n\ge\ 2\ $ (not just for $\ n=2).\ $ This follows from the Borsuk's theorem for each $\ n\ge 2\ $ just as it does for its special case of $\ n=2$.

I like my (previous) answer in its simple form. Thus I will expand it here (it should not be a waste, I hope), and first of all I will fulfill Todd's request, see Thm 1 below: we want to show that a closed subset of $\ \mathbb R^2,\ $ which separates $\ \mathbb R^2,\ $ is not totally disconnected. The proof is smoother in the compact case, but in general it needs just a minor extra consideration.

A direct dimension-flavor approach is possible. However I feel like applying the elegant Borsuk's theorem based on his separation criterion. Eilenberg and Steenrod cover this material beautifully in their Chapter 11 (it's like a perfectly elementary appendix) to their "Foundations of Algebraic Topology".

BORSUK's THEOREM   Let $\ X\ $ be a closed subset of $\ S^n.\ $ Then $\ S^n\setminus X\ $ is disconnected $\ \Leftrightarrow\ $ there exists a continuous map $\ f:X\rightarrow S^{n-1}\ $ which is not homotopic to a constant.

On this occasion I also have selected this direction of presentation because Bill Thurston in a respective post wrote about the Alexander duality in a way which harmonized with the above. I read that Bill's post only after Kristal's answer (which didn't really answer) since it referenced to the said Bill's post; the very first comment of this whole thread, just under the QUESTION of the threat, written by me before Kristal's answer, read:

    It is not possible. Where does this problem come from?

Now, in the compact context the being totally disconnected is equivalent to being $0$-dimensional. This means for the totally disconnected (i.e. $0$-dimensional) closed set $\ X\subseteq S^n\ $ every continuous map $\ f:X\rightarrow S^{n-1}\ $ is homotopically trivial for $\ n\ge 2.\ $ Thus for a compact $\ X\subseteq\mathbb R^2\ $ which disconnects $\ R^2\ $ the requested theorem already holds, i.e. $\ X\ $ is not totally disconnected (add $\infty$ to the unbounded component of $\ \mathbb R^2\setminus X).\ $ Thus we have to include in the proof only the (nuisance :-) non-compact case (while the theorem is formulated for both):

THEOREM 1   Let $\ X\ $ be a closed subset of $\ \mathbb R^2\ $ such that $\ \mathbb R^2\setminus X\ $ is not connected. Then $\ X\ $ is not totally disconnected.

PROOF The case of compact $\ X\ $ was covered above. Now let $\ X\ $ be closed and not compact. Then $\ Y := X\cup\{\infty\}\ $ is compact. Of course $\ Y\ $ disconnects $\ \{\infty\}\cup \mathbb R^2 = S^2.\ $ Thus by the Borsuk's theorem and the remarks which followed, $\ \dim(Y)\ge 1.\ $ However $\ X\ $ is $\sigma$-compact, $\ X=\bigcup_{n=1\ 2\ \ldots}X_n\ $ where each $\ X_n\ $ is compact. Thus $\ \dim(\{\infty\}\cup\bigcup_{n=1\ 2\ \ldots}X_n) = \dim Y \ge 1,\ $ and $\ \dim(\{\infty\}) = 0.\ $ Thus $\ \exists_n \dim(X_n)\ge 1.\ $ Such $\ X_n\ $ is not totally disconnected, hence neither is $\ X$.   END of Proof

Todd's request's fullfilled, I will continue with the rest later.

I like my (previous) answer in its simple form. Thus I will expand it here (it should not be a waste, I hope), and first of all I will fulfill Todd's request, see Thm 1 below: we want to show that a closed subset of $\ \mathbb R^2,\ $ which separates $\ \mathbb R^2,\ $ is not totally disconnected. The proof is smoother in the compact case, but in general it needs just a minor extra consideration.

A direct dimension-flavor approach is possible. However I feel like applying the elegant Borsuk's theorem based on his separation criterion. Eilenberg and Steenrod cover this material beautifully in their Chapter 11 (it's like a perfectly elementary appendix) to their "Foundations of Algebraic Topology".

BORSUK's THEOREM   Let $\ X\ $ be a closed subset of $\ S^n.\ $ Then $\ S^n\setminus X\ $ is disconnected $\ \Leftrightarrow\ $ there exists a continuous map $\ f:X\rightarrow S^{n-1}\ $ which is not homotopic to a constant.

On this occasion I also have selected this direction of presentation because Bill Thurston in a respective post wrote about the Alexander duality in a way which harmonized with the above. I read that Bill's post only after Kristal's answer (which didn't really answer) since it referenced to the said Bill's post; the very first comment of this whole thread, just under the QUESTION of the threat, written by me before Kristal's answer, read:

    It is not possible. Where does this problem come from?

Now, in the compact context the being totally disconnected is equivalent to being $0$-dimensional. This means for the totally disconnected (i.e. $0$-dimensional) closed set $\ X\subseteq S^n\ $ every continuous map $\ f:X\rightarrow S^{n-1}\ $ is homotopically trivial for $\ n\ge 2.\ $ Thus for a compact $\ X\subseteq\mathbb R^2\ $ which disconnects $\ R^2\ $ the requested theorem already holds, i.e. $\ X\ $ is not totally disconnected (add $\infty$ to the unbounded component of $\ \mathbb R^2\setminus X).\ $ Thus we have to include in the proof only the (nuisance :-) non-compact case (while the theorem is formulated for both):

THEOREM 1   Let $\ X\ $ be a closed subset of $\ \mathbb R^2\ $ such that $\ \mathbb R^2\setminus X\ $ is not connected. Then $\ X\ $ is not totally disconnected.

PROOF The case of compact $\ X\ $ was covered above. Now let $\ X\ $ be closed and not compact. Then $\ Y := X\cup\{\infty\}\ $ is compact. Of course $\ Y\ $ disconnects $\ \{\infty\}\cup \mathbb R^2 = S^2.\ $ Thus by the Borsuk's theorem and the remarks which followed, $\ \dim(Y)\ge 1.\ $ However $\ X\ $ is $\sigma$-compact, $\ X=\bigcup_{n=1\ 2\ \ldots}X_n\ $ where each $\ X_n\ $ is compact. Thus $\ \dim(\{\infty\}\cup\bigcup_{n=1\ 2\ \ldots}X_n) = \dim Y \ge 1,\ $ and $\ \dim(\{\infty\}) = 0.\ $ Thus $\ \exists_n \dim(X_n)\ge 1.\ $ Such $\ X_n\ $ is not totally disconnected, hence neither is $\ X$.   END of Proof

Todd's request's fulfilled; now back to the generalization, while George Lowther has already provided a still more general result (nice!)

The proof in my first answer used only the assumption that set $\ A\ $ was disconnected, and not more--it didn't mention that $\ A\ $ was totally disconnected. Thus (as promised in my first answer) I really proved:

THEOREM 2   If $\ \mathbb R^2 = A\cup B\ $, and $\ A\ $ is not connected, then $\ B\ $ is not totally disconnected.

In other words:

THEOREM 2'   If $\ \mathbb R^2 = A\cup B\ $, and $\ B\ $ is totally disconnected then $\ A\ $ is connected.

Actually, all this holds for $\ \mathbb R^n\ $ for every $\ n\ge\ 2 (not just for $\ n=2). This follows from the Borsuk's theorem for each $\ n\ge 2\ $ just as it does from its special case of $\ n=2$.

I like my (previous) answer in its simple form. Thus I will expand it here (it should not be a waste, I hope), and first of all I will fulfill Todd's request, see Thm 1 below: we want to show that a closed subset of $\ \mathbb R^2,\ $ which separates $\ \mathbb R^2,\ $ is not totally disconnected. The proof is smoother in the compact case, but in general it needs just a minor extra consideration.

A direct dimension-flavor approach is possible. However I feel like applying the elegant Borsuk's theorem based on his separation criterion. Eilenberg and Steenrod cover this material beautifully in their Chapter 11 (it's like a perfectly elementary appendix) to their "Foundations of Algebraic Topology".

BORSUK's THEOREM   Let $\ X\ $ be a closed subset of $\ S^n.\ $ Then $\ S^n\setminus X\ $ is disconnected $\ \Leftrightarrow\ $ there exists a continuous map $\ f:X\rightarrow S^{n-1}\ $ which is not homotopic to a constant.

On this occasion I also have selected this direction of presentation because Bill Thurston in a respective post wrote about the Alexander duality in a way which harmonized with the above. I read that Bill's post only after Kristal's answer (which didn't really answer) since it referenced to the said Bill's post; the very first comment of this whole thread, just under the QUESTION of the threat, written by me, read:

    It is not possible. Where does this problem come from?

Now, in the compact context the being totally disconnected is equivalent to being $0$-dimensional. This means for the totally disconnected (i.e. $0$-dimensional) closed set $\ X\subseteq S^n\ $ every continuous map $\ f:X\rightarrow S^{n-1}\ $ is homotopically trivial for $\ n\ge 2.\ $ Thus for a compact $\ X\subseteq\mathbb R^2\ $ which disconnects $\ R^2\ $ the requested theorem already holds, i.e. $\ X\ $ is not totally disconnected (add $\infty$ to the unbounded component of $\ \mathbb R^2\setminus X).\ $ Thus we have to include in the proof only the (nuisance :-) non-compact case (while the theorem is formulated for both):

THEOREM 1   Let $\ X\ $ be a closed subset of $\ \mathbb R^2\ $ such that $\ \mathbb R^2\setminus X\ $ is not connected. Then $\ X\ $ is not totally disconnected.

PROOF The case of compact $\ X\ $ was covered above. Now let $\ X\ $ be closed and not compact. Then $\ Y := X\cup\{\infty\}\ $ is compact. Of course $\ Y\ $ disconnects $\ \{\infty\}\cup \mathbb R^2 = S^2.\ $ Thus by the Borsuk's theorem and the remarks which followed, $\ \dim(Y)\ge 1.\ $ However $\ X\ $ is $\sigma$-compact, $\ X=\bigcup_{n=1\ 2\ \ldots}X_n\ $ where each $\ X_n\ $ is compact. Thus $\ \dim(\{\infty\}\cup\bigcup_{n=1\ 2\ \ldots}X_n) = \dim Y \ge 1,\ $ and $\ \dim(\{\infty\}) = 0.\ $ Thus $\ \exists_n \dim(X_n)\ge 1.\ $ Such $\ X_n\ $ is not totally disconnected, hence neither is $\ X$.   END of Proof

Todd's request's fullfilled, I will continue with the rest later.

I like my (previous) answer in its simple form. Thus I will expand it here (it should not be a waste, I hope), and first of all I will fulfill Todd's request, see Thm 1 below: we want to show that a closed subset of $\ \mathbb R^2,\ $ which separates $\ \mathbb R^2,\ $ is not totally disconnected. The proof is smoother in the compact case, but in general it needs just a minor extra consideration.

A direct dimension-flavor approach is possible. However I feel like applying the elegant Borsuk's theorem based on his separation criterion. Eilenberg and Steenrod cover this material beautifully in their Chapter 11 (it's like a perfectly elementary appendix) to their "Foundations of Algebraic Topology".

BORSUK's THEOREM   Let $\ X\ $ be a closed subset of $\ S^n.\ $ Then $\ S^n\setminus X\ $ is disconnected $\ \Leftrightarrow\ $ there exists a continuous map $\ f:X\rightarrow S^{n-1}\ $ which is not homotopic to a constant.

On this occasion I also have selected this direction of presentation because Bill Thurston in a respective post wrote about the Alexander duality in a way which harmonized with the above. I read that Bill's post only after Kristal's answer (which didn't really answer) since it referenced to the said Bill's post; the very first comment of this whole thread, just under the QUESTION of the threat, written by me before Kristal's answer, read:

    It is not possible. Where does this problem come from?

Now, in the compact context the being totally disconnected is equivalent to being $0$-dimensional. This means for the totally disconnected (i.e. $0$-dimensional) closed set $\ X\subseteq S^n\ $ every continuous map $\ f:X\rightarrow S^{n-1}\ $ is homotopically trivial for $\ n\ge 2.\ $ Thus for a compact $\ X\subseteq\mathbb R^2\ $ which disconnects $\ R^2\ $ the requested theorem already holds, i.e. $\ X\ $ is not totally disconnected (add $\infty$ to the unbounded component of $\ \mathbb R^2\setminus X).\ $ Thus we have to include in the proof only the (nuisance :-) non-compact case (while the theorem is formulated for both):

THEOREM 1   Let $\ X\ $ be a closed subset of $\ \mathbb R^2\ $ such that $\ \mathbb R^2\setminus X\ $ is not connected. Then $\ X\ $ is not totally disconnected.

PROOF The case of compact $\ X\ $ was covered above. Now let $\ X\ $ be closed and not compact. Then $\ Y := X\cup\{\infty\}\ $ is compact. Of course $\ Y\ $ disconnects $\ \{\infty\}\cup \mathbb R^2 = S^2.\ $ Thus by the Borsuk's theorem and the remarks which followed, $\ \dim(Y)\ge 1.\ $ However $\ X\ $ is $\sigma$-compact, $\ X=\bigcup_{n=1\ 2\ \ldots}X_n\ $ where each $\ X_n\ $ is compact. Thus $\ \dim(\{\infty\}\cup\bigcup_{n=1\ 2\ \ldots}X_n) = \dim Y \ge 1,\ $ and $\ \dim(\{\infty\}) = 0.\ $ Thus $\ \exists_n \dim(X_n)\ge 1.\ $ Such $\ X_n\ $ is not totally disconnected, hence neither is $\ X$.   END of Proof

Todd's request's fullfilled, I will continue with the rest later.