Roughly, a compact, Hausdorff space $X$ has covering dimension $\leqslant n$ if each finite cover $\mathcal{U}$ of $X$ can be refined by a cover $\mathcal{V}$ such that each point $x\in K$ belongs to at most $n+1$ sets from the new, refined cover.
Hence my question:
Suppose that $X$ is one-dimensional compact Hausdorff space and let $\mathcal{U}$ be a finite open cover for $X$. Can $\mathcal{U}$ be refined in such a way that each element $x\in X$ belongs to precisely two sets from the new cover?
This doesn't appear to be true for even the unit interval $[0,1]$. Take the simple open cover $\mathcal{U} = \{ [0,\frac{2}{3} ) , (\frac{1}{3},1] \}$, and suppose $\mathcal{V}$ is an open refinement of $\mathcal{U}$ in which all points belong to exactly two sets.
Let $V_1,V_2$ be the two sets in $\mathcal{V}$ containing $\frac{1}{2}$. Note that either $\alpha = \sup ( V_1 \cap V_2 ) < 1$ or $\beta = \inf ( V_1 \cap V_2 ) > 0$, so without loss of generality assume the latter holds. Then $\beta$ does not belong to at least one of $V_1$ or $V_2$, so there must be a $W \in \mathcal{V}$ containing $\beta$ which is different from $V_1 , V_2$. But it follows that $W \cap V_1 \cap V_2 \neq \emptyset$, contradicting our assumption on $\mathcal{V}$.