Does every connected topological space have the property that you can walk around a finite open cover to get from any point to any other? Let $U$ denote a finite open cover of the unit interval $[0,1]$ indexed by a set $I$. We get a graph as follows:


*

*The vertexes are the element of $I$.

*There's an edge from $i$ to $j$ if and only if $U_i \cap U_j$ share an element.


Lets call this graph $I$, since it's basically just the index set, but with a bit more structure.
There's two sets of vertexes of particular interest:


*

*$I_0 = \{i \in I : U_i \ni 0\}.$

*$I_1 = \{i \in I : U_i \ni 1\}.$


So $(I,I_0,I_1)$ is a graph with two distinguished sets of vertexes. We can think of these as 'start nodes' and 'end nodes.'
It seems to be the case that every graph-with-startnodes-and-endnodes obtained in this way has the property that there exists a walk from a startnode to an endnode. I don't have a firm understanding of the topological reasons why this is true; I can't prove it. However it seems to be largely a consequence of the connectedness of $[0,1]$. We might therefore conjecture the following:

Conjecture. Let $X$ denote a connected topological space and $U$ denote a finite open cover of $X$ indexed by a set $I$. Consider the graph $I$ defined by the sharing of elements (as above). Then $I$ is connected as a graph, meaning there's a walk between any two vertexes.

I'd like to know:

Questions. Firstly, is this true?
If not, then secondly, what further assumptions do we need on $X$ before it becomes true?


The claim is true for connected spaces.
Assume by contradiction that your graph is disconnected. Write $G=G_1 \cup G_2$ as the union of two disjoint subgraphs $G_1=(V_1, E_1), G_2(V_2,E_2)$. 
Define 
$$W_1= \bigcup_{j \in V_1}  U_{j} \\
W_2= \bigcup_{j \in V_2}  U_{j} $$
Then $W_1, W_2$ are open and $X=W_1 \cup W_2$.
Now, since there is no edge between $E_1$ and $E_2$ you have $W_1 \cap W_2 =\emptyset$.
This gives that $W_1, W_2$ are clopen in $X$ and hence $X$ is disconnected. 
