Say, we define the Lebesgue covering dimension (LCD) like this:
A set $S\in \mathbb R^n$ has LCD $d\in \mathbb N$ if and only if $d$ is the smallest natural number such that for any open cover $(C_i)_{i\in I}$ of $S$ there exists a refinement $(C_j)_{j\in J}$ such that for any $x \in S$: $x$ lies in at most $d+1$ sets of the refinement. (If there is no such $d\in \mathbb N$, we say that $\dim_\text{LCD}S=\infty$.)
I wanted to show that $\dim_\text{LCD}[0,1]=1$.
The geometry of it is quite clear if one draws the picture, and one can convince himself by drawing a cover and then a refinement. Yet I'm not quite sure how to argue properly.
For example, if $S$ is totally disconnected, then it's easy to conclude that its LCD is $0$. Would the fact that $[0,1]$ is connected be helpful for showing $\dim_\text{LCD}[0,1]=1$? Or can we prove it straight from the definition?
The connectedness of $[0,1]$ implies that the Lebesgue covering dimension is at least $1$, since an open cover such that every point lies in only one member of the cover must consist of clopen sets, hence by connectedness be the trivial cover $\{[0,1]\}$.
To show that the dimension is at most $1$, consider an open cover $\mathfrak{U}$ of $[0,1]$. Since $[0,1]$ is compact, there is an $\varepsilon > 0$ such that every subset of $[0,1]$ whose diameter is at most $\varepsilon$ is contained in one of the members of $\mathfrak{U}$ (the cover $\mathfrak{U}$ has positive Lebesgue number). Then
$$\mathfrak{V} = \left\{ \left((2k+1)\frac{\varepsilon}{3} - \frac{\varepsilon}{2}, (2k+1)\frac{\varepsilon}{3}+\frac{\varepsilon}{2}\right) : 0 \leqslant k \leqslant \frac{1}{2}\left(\frac{3}{\varepsilon}-1\right)\right\}$$
is a refinement of $\mathfrak{U}$ such that no point of $[0,1]$ lies in more than $2$ covering sets, hence $\dim_{\text{LCD}} [0,1] \leqslant (2-1)$.