Given connected topological spaces $X_i$ and a totally disconnected space $Y$, is there a unique topological space $X$ with components homeomorphic to $X_i$ and $X/\sim$ homeomorphic to $Y$? ($\sim$ is the partition of components.) The question may be sub-divided into existence and uniqueness.
This may be obviously true to a topologist, but since topology is full of surprises, I thought I'd check with the experts.
(Second quick question: what is a "minimal" example of a non-metrizable compact Hausdorff space? "Minimal" must include separable first countable and possibly other nice properties such as connected or totally disconnected.)
Let $Y$ be a totally disconnected space with topology $\tau_Y$, and for each $y\in Y$ let $X_y$ be a connected space with topology $\tau_y$. Without loss of generality assume that the spaces $X_y$ are pairwise disjoint, and let $X=\bigcup_{y\in Y}X_y$. For each $y\in Y$ fix a point $p_y\in X_y$, and let $\tau_y^*=\{U\in\tau_y:p_y\notin U\}$. If $y\in U\in\tau_Y$ and $p_y\in V\in\tau_y$ let
$$B_y(U,V)=V\cup\bigcup_{z\in U\setminus\{y\}}X_z\;,$$
and let $\mathscr{B}_y$ be the collection of all such sets $B_y(U,V)$. Finally, let $\tau$ be the topology on $X$ generated by the base
$$\bigcup_{y\in Y}(\mathscr{B}_y\cup\tau_y^*)\;.$$
It’s not hard to check that each $X_y$ inherits the topology $\tau_y$ from $\tau$ and is a component of $X$, and that the obvious map from $X$ to $Y$ is a quotient map: for any $A\subseteq Y$ we clearly have $\bigcup_{y\in A}X_y\in\tau$ iff $A\in\tau_Y$.
However, this $X$ is not necessarily unique. For a simple counterexample let $S=\{2^{-n}:n\in\Bbb N\}$ and $Y=\{0\}\cup S$. For $n\in\Bbb N$ we want the component of $X$ corresponding to $2^{-n}$ to be a singleton, and the component corresponding to $0$ is to be a copy of the unit interval $[0,1]$. Let
$$X_0=\big(\{0\}\times[0,1]\big)\cup\big(S\times\{0\}\big)$$
and
$$X_1=\big(\{0\}\times[-1,1]\big)\cup\big(S\times\{0\}\big)\;;$$
in each case the projection to the first coordinate is a quotient map onto $Y$. Note that in both $X_0$ and $X_1$ the origin is the unique limit point of the set of isolated points, so any homeomorphism of $X_0$ and $X_1$ would have to send it to itself. However, removing it from $X_0$ leaves a space with one component that is not a singleton, while removing it from $X_1$ leaves a space with two such components. Thus, $X_0$ and $X_1$ are not homeomorphic.