Show that the Lebesgue Covering Dimension of the unit square $I^2$ with $I=[0,1]$ is two.
I know that a compact subset of Euclidean space $\Bbb R^n$ has Lebesgue dimension at most $n$. So it suffices to show that $\dim I^2\geq 2$. I first demonstrate an open covering of order three.
Let $G=[0, 1/2)\cup(1/2, 1]$. Then if $\pi_1$ and $\pi_2$ are the projection maps of $I^2$ onto $I$, consider the collection $\mathcal{A}=\{(0,1)^2, \pi_1^{-1}(G), \pi_2^{-1}(G)\}$. $\mathcal{A}$ is clearly an open covering and has order three.
Now let $\mathcal{B}$ be an open refinement of $\mathcal{A}$ that still covers $I^2$. I want to show that $\mathcal{B}$ also has order three. I know that $\mathcal{B}$ must contain three distinct elements since it is a cover of $I^2$, a refinement of $\mathcal{A}$, and each element of $\mathcal{A}$ has an element which the other two do not.
I know that $\mathcal{B}$ cannot have order one. If it did, and $U\in\mathcal{B}$ and $V$ were the union of the rest of the elements of $\mathcal{B}$ then $(U,V)$ would form a separation of $I^2$ contradicting its connectedness.
Now I'm stuck on showing that $\mathcal{B}$ cannot have order two. I know that $\mathcal{B}$ has a Lebesgue number $\delta$, but it hasn't helped much. I thought about projecting the elements of $\mathcal{B}$ onto $I$ which I know has dimension one, but this doesn't seem to lead anywhere. I've tried working with partitions of unity, but they also haven't lent any insight.
Any help is appreciated.
UPDATE
At Mirko's suggestion, I tried a different beginning open covering with order three. I think I've simplified the problem somewhat, but I feel like the gap required to finish this proof is still left to be traversed.
Let's start with the open covering Mirko suggested: $$\mathcal{A}=\{[0,2/3)\times I, (1/3, 1]\times [0,2/3), (1/3,1]\times(1/3,1]\}$$ $\mathcal{A}$ clearly has order three. Now let $\mathcal{B}$ be an open refinement of $\mathcal{A}$ which still covers $I^2$. By similar reasoning as above, $\mathcal{B}$ cannot have order one since $I^2$ is connected. So we just need to show that $\mathcal{B}$ does not have order two. Suppose it did to (hopefully eventually) get a contradiction. Let $\mathcal{B}'$ denote the set of all subsets of $I^2$ which occur as a connected component of some element of $\mathcal{B}$. Since $I^2$ is locally connected, $\mathcal{B}'$ is still an open covering of $I^2$. Further, $\mathcal{B}'$ still refines $\mathcal{A}$ and also has order two (assuming that $\mathcal{B}$ does). Now since $I^2$ is compact $\mathcal{B}'$ contains a finite subcover $\mathcal{C}$ of $I^2$. We can assume that removing any one element from $\mathcal{C}$ results in a collection which does not cover $I^2$. $\mathcal{C}$ still refines $\mathcal{A}$ and has order two (assuming $\mathcal{B}$ does).
We now have a finite collection $\mathcal{C}$ consisting of connected open sets (hence also path-connected) refining $\mathcal{A}$ such that any subcollection of $\mathcal{C}$ is not a cover of $I^2$, which, overmore, supposedly has order two. If we can show this is impossible, we have found our contradiction.
If we let $\mathcal{A}=\{A_1, A_2, A_3\}$ (where the $A_i$ represent the respective sets already stated), then each element of $\mathcal{C}$ lies entirely in one of the $A_i$ (perhaps two of the $A_i$) since each element of $\mathcal{C}$ and $\mathcal{A}$ is connected. From here I don't see what to try next.