Quotient of plane set embeds in $\mathbb R ^3$? Let $X\subseteq \mathbb R ^2$ be a compact subspace of the plane.  Let $A\subseteq X$ be closed.  It is well-known that the quotient $X/A$ is metric and separable, and therefore embeds into the Hilbert space $\mathbb R^\omega$. Does it embed into $\mathbb R ^3$?  It seems like the answer should be yes.
EDIT: The Menger-Nöbeling theorem (1932) states that if X is compact metric separable and of dimension n, then it embeds as a subspace of Euclidean space of dimension 2n + 1.  So the answer to my question is yes if $X$ is a 1-dimensional.  But of course $X$ could have dimension $2$...
 A: The answer is indeed yes. Because this sort of thing is usually a little
easier to see with compact spaces I will first show that if $A$ is a
closed subset of $S^n$, then $S^n/A$ can be embedded in $\mathbb{R}^{n+1}$.
(To get the basic intuition behind the proof, ask yourself what might be the easiest way of bringing two points on an inflated balloon together.)
Given a nonempty closed set $A \subset S^n \subset \mathbb{R}^{n+1}$, we define a
continuous function $d_A: \mathbb{R}^{n+1} \to [0, +\infty)$ by
$d_A(x) = \inf \{ \| x - y \| \mid y \in A  \}$. Clearly then $f_A: x \mapsto d_A(x)\, x$ defines a continuous map $S^n \to \mathbb{R}^{n+1}$ 
and it is easy to verify that the nonempty fibres of $f$ are $A$ and the singletons in $S^n \setminus A$. Since $f_A$ is a closed map by compactness of $S^n$,
this means that there is a decomposition $f_A = i \circ q$ where 
$q: S^n \to S^n/A$ is the canonical map and 
$i: S^n/A \to \mathbb{R}^{n+1}$ is an embedding. $\square$
Now if $X$ is a subspace of $S^n$ and $A$ is a nonempty compact subset of $X$,
then $A$ is closed in $S^n$ and  we have $f_A = i \circ q$ as above. The 
restriction $i|_{q[X]}$ is still an embedding and 
$q|_{q^{-1}[q[X]]}^{q[X]}$ is still
a quotient map. Since $A \subset X$ we have $q^{-1}[q[X]] = X$, hence
$q|_X^{q[X]}$ is a quotient map from $X$ to $X/A$. Thus we can conclude that if $X$ is any space embeddable in $S^n$ and $A \subset X$ is compact,
then $X/A$ is embeddable in $\mathbb{R}^{n+1}$.
A: I think the answer is no, in general.  The Klein bottle is a quotient of $[0,1]^2$ but is not embeddable in $\mathbb R ^3$.
EDIT: Actually, this is not a counterexample.  I was interested in shrinking the entire closed set to a single point.  
