The Premise:
Let $X$ be a topological space. For any pair of points $x_0,x_1\in X$, let $\mathrm N_X(x_0,x_1)$ be the set of all subspaces $S$ of $X$ such that $x_0,x_1\in S$ and $S\cong I$ for some interval $I\subseteq\Bbb R$.
We say that $X$ is "noodly" if and only if
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$X$ is second countable
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$X$ is Hausdorff
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$X$ is locally metrizable
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$X$ is path connected and contains at least two distinct points (equivalent: $X$ is arc-connected, $\mathrm N_X(x_0,x_1)\ne\emptyset$ for all $x_0,x_1\in X$)
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$\bigcup_{x\in X}\bigcup\mathrm N_X(x_0,x)=X$ for all $x_0\in X$
Every connected topological manifold is trivially noodly, and gluing along noodly subspaces preserves noodliness. The gluing of disjoint (though not necessarily distinct) manifolds provides the motivating example for noodly spaces (also homotopy stuff, because homotopy is only intuitive for noodly spaces anyway.)
For a noodly space $X$, a "decomposition" of $X$ is a [countable?] set $\mathcal A$ of pairwise disjoint noodly spaces and an equivalence relation $E$ on $\bigcup\mathcal A$ such that $\bigcup \mathcal A/E\cong X$.
The Question:
Are there any noodly spaces which cannot be decomposed into a set of pairwise disjoint manifolds?
That is, are there any noodly spaces $X$ for which there does not exist a set $\mathcal M$ of pairwise disjoint manifolds such that $X\cong\bigcup\mathcal M/E$ for some equivalence relation $E$?
Best Answer
Your condition 5 follows from path connectedness, at least under Hausdorff, and thus is redundant.
Therefore spaces that you consider are precisely: second countable, locally metrizable, Hausdorff and path connected. And you ask: is such space a quotient of a (possibly disconnected) manifold? The answer is: not necessarily. Note that manifolds are locally connected, and thus so are their quotients. But there's a simple example of a space that satisfies all those conditions, and additionally is not locally connected: the comb space.