Let $\mathcal{F}$ be this partition. As noted above it must be uncountable. We may as well assume it lives on the interval $(0,1)$ and add the singleton sets $ \lbrace 0\rbrace $ and $\lbrace 1\rbrace$ to it to obtain a partition of $[0,1]$. Now $\mathcal{F}$ is an uncountable subset of the space of closed subsets of $[0,1]$ endowed with the Hausdorff metric. The latter space is separable metric, so $\mathcal{F}$ contains a sequence $\langle F_n:n\in\mathbb{N}\rangle$ that converges to a point $F$ of $\mathcal{F}$. The union $F\cup\bigcup_n F_n$ is closed: if $x$ is outside the union, in particular outside $F$, let $\epsilon=\frac12 d(x,F)$. Then there is an $N$ such that $F_n\subseteq B(F, \epsilon)$ for $n\ge N$. This now easily implies that $x$ is not in the closure of the union.

Let $\mathcal{F}$ be this partition. As noted above it must be uncountable. We may as well assume it lives on the interval $(0,1)$ and add the set $ \lbrace 0, 1\rbrace$ to each member to obtain a family of closed subsets of $[0,1]$. Now $\mathcal{F}$ is an uncountable subset of the space of closed subsets of $[0,1]$ endowed with the Hausdorff metric. The latter space is separable metric, so $\mathcal{F}$ contains a non-trivial sequence $\langle F_n:n\in\mathbb{N}\rangle$ that converges to a point $F$ of $\mathcal{F}$. The union $F\cup\bigcup_n F_n$ is closed: if $x$ is outside the union, in particular outside $F$, let $\epsilon=\frac12 d(x,F)$. Then there is an $N$ such that $F_n\subseteq B(F, \epsilon)$ for $n\ge N$. This now easily implies that $x$ is not in the closure of the union.