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For any cover $\cal{U}$ of a topological space $X$ there is a finest partition $P(\cal{U})$ refined by $\cal{U}$. Each element of $\cal{U}$ is contained in a unique element of $P(\cal{U})$. Willard, General topology, Ex. 36A. I ask for a proof please. Also if we take the cover $[0,1/3]\cup[1/4,2/3]\cup[1/2,1]$ of $[0,1]$ which is the finest partition of the proposition?

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For your specific part: suppose that $\mathcal{P}$ is a partition of $[0,1]$ that refines your cover $\{[0,\frac13], [\frac14,\frac23],[\frac12,1]\}$.

Then some $P_1 \in \mathcal{P}$ satisfies $[0,\frac13] \subseteq P_1$ and also some $P_2 \in \mathcal{P}$ exists such that $[\frac14,\frac23]\subseteq P_2$. But $\frac14$ lies in both and so this forces $P_1=P_2$ (two elements of a partition are disjoint or equal). Similarly, a $P_3$ containing $[\frac12,1]$ must equal $P_2$ (and thus $P_1$!) as $\frac12$ is in both. It follows that $P_1=P_2=P_3$ and the partition $\mathcal{P}$ equals the quite boring $\{[0,1]\}$.

This gives an idea for a partition (which is really the set of classes of an equivalence relation) for refining any $\mathcal{U}$:

Define $x \sim y$ iff there is a "path in $\mathcal{U}$ from $x$ to $y$":

$$x \sim y \iff \exists n\ge 1, \exists U_1, \ldots, U_n \in \mathcal{U}: (x \in U_1) \land (y \in U_n) \land (\forall 1 \le i \le n-1: U_i \cap U_{i+1} \neq \emptyset\tag{1}$$

and let $\mathcal{P}$ be the set of equivalence classes. It's indeed easy to check that this defines an equivalence relation on $X$ and that the classes refine the members of $\mathcal{U}$ (if $x,y \in U \in \mathcal{U}$ then $x \sim y$) and is as required.

Bonus theorem: $X$ is connected iff for every open cover $\mathcal{U}$ of $X$ and every $x,y \in X$ there is a path in $\mathcal{U}$ from $x$ to $y$. This is quite handy in some proofs involving local compactness, or local (path-)connectedness etc.

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