Consider $\overline{B_{\delta}(\vec{x})}\subseteq\mathbb{R}^{n}$ (a closed ball of radius $\delta>0$) Using Heine-Borel I can create a finite subcover for every open cover call it $$O: = \bigcup_{i={1}}^{n} U_i$$
It seems to me that there is always a $1 \leq j\leq n$ such that for a fixed i, $U_{i}\cap U_{j}\neq \emptyset$ for $i\neq j$. If this is true then I can always order the union so that $U_{i+1}\cap U_{i}\neq \emptyset$. Can this be done?
When you get down to it, this is actually a combinatoric graphing problem and an obviously false one.
Let's have an open cover consisting of $U_1, .... U_n$ and you have determined that for every $U_i$ there is at least one other $U_j$ where $U_i$ and $U_j$ intersect.
Now lets make a graph of vertices $x_i$ were $x_i$ represents the set $U_i$ and let's create an edge connecting $x_i$ to $x_j$ if $U_i$ intersects $U_j$. In the resulting graph every vertex has at least one edge to another, and the graph is connected.
Now you are asking if it will always be possible to find a path going through each vertex exactly once.
The answer to that is obviously no.
Consider a simple trifold. A single center point connected to three otherwise unconnected points. A,B,C are each connected to X and those are the only connection. You can't travel passing through each point once because to get from $j \in \{ A,B,C\}$ to $k \in \{A,B,C\}$ you must pass through X. And to get from the from $k$ to $m \in \{A,B,C\}$, you must pass through X again.
So consider $B$ equals the unit Ball. $X$ is the entire plane except for three distinct points on the interior of $B$. Let $J$, $K$, $L$ be small open neighborhoods of those three points that are missing for $X$ so that $J$ and $K$ and $L$ are mutually distinct.
Then $B \subset J \cup K \cup L \cup X$ is an open cover. $J\cap X \ne \emptyset$ and $K\cap X \ne \emptyset$ and $L\cap X \ne \emptyset$ and $X\cap J \ne \emptyset$. But we can not reorder them as $U_1, U_2, U_3,U_4$ so that $U_1 \cap U_2 \ne \emptyset$ and $U_2 \cap U_3\ne \emptyset$ and $U_3 \cap U_4\ne \emptyset$ because $X$ would need to be the common set in each pair and there is no common set in each pair.