A $\sigma$-compact separable zero-dimensional metric space has property stronger than $\sigma$-compactness.

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If $X$ is a $\sigma$-compact, zero-dimensional, separable metric space, how can I prove that there exists a sequence $(A_n)_n$ of compact sets such that $\bigcup_{n=1}^\infty A_n = X$, $A_i\cap A_j = \emptyset$ for $i\neq j$ (so a partition of the space $X$) and $\text{diam}(A_n)\to 0$?

I thought about obtaining a countable basis $\mathcal{B} = \bigcup_{n=1}^\infty \mathcal{B}_n$ of clopen sets such that $\mathcal{B}_n$ is a disjoint cover of $X$ consisting of sets of diameter $\leq \frac{1}{n}$, and $\mathcal{B}_{n+1}$ is a refinement of $\mathcal{B}_n$.

But I am stuck in that I can't really take a compact subset of $X$ and cover it with those sets arbitrarily, there could be a part that "sticks out". How do prove this then? I am completely stuck.