I was looking for a related fact, and surprisingly couldn't find anything relevant, except of this question. Even though it was answered 10 years ago, perhaps the following result could be useful to somebody.

**Proposition.** Let $X$ be a connected metric space that contains more than one point. Let $A\subset X$ be totally disconnected and locally compact with respect to the subspace topology. Then $A$ is nowhere dense.

**Proof.** First, let us show $int A =\varnothing$. Assume that $U$ is an open set in $X$ such that $\overline{U}$ is compact and contained in $A$, and $x\in U$. Since a totally disconnected locally compact paracompact space is zero-dimensional (see 6.2.9 in Engelking's General Topology), there is an open neighborhood $V\subset U$ of $x$ that is clopen in $A$. Then, there are closed set $F$ in $X$ and open set $W$ in $X$ such that $V=A\cap F=A\cap W$. Since $V\subset U\subset A$, $V=U\cap W$ is open in $X$. Since $V\subset \overline{U}\subset A$, $V=\overline{U}\cap F$ is closed in $X$. Hence, $V$ is nonempty and clopen in $X$ which contradicts its connectedness.

Now recall that a locally compact set is open in its closure (see 3.3.9 in Engelking). Hence, $A=\overline{A}\cap U$, for some open $U\subset X$. Assume $int \overline{A}\ne\varnothing$. Then, there is $x\in int \overline{A} \cap A= int \overline{A} \cap\overline{A}\cap U=int \overline{A} \cap U\subset \overline{A} \cap U=A$, from where $x\in int \overline{A} \cap U\subset A$, and so $x\in int A$. Contradiction with the previous step.

**Corollary.** If $X$ is a connected complete metric space that contains more than one point, it cannot be covered by a countable collection of totally disconnected locally compact subsets.

**In particular**, we cannot remove a closed totally disconnected set (e.g. a single point) from a continuum to make it totally disconnected.

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