Here is the theorem 1-4:
If B is compact and $\mathcal{O}$ is an open cover of $\{x\}\times B$, then there is an open set U $\in R^n$ containing $x$ such that $U\times B$ is covered by a finite number of sets in $\mathcal{O}$.
Spivak's proof goes like this:
Since $\{x\} \times B$ is compact, we can assume at the outset that $\mathcal{O}$ is finite, and we need only find the open set $U$ such that $U\times B$ is covered by $\mathcal{O}$. For each $y \in B$ the point $(x,y)$ is in some open set $W$ in $\mathcal{O}$. Since $W$ is open, we have $(x,y) \in U_y \times V_y \subset W$ for some open rectangle $U_y \times V_y$.
The sets $V_y$ cover the compact set $B$, so a finite number $\{V_{y_1}, ... , V_{y_k}\}$ also cover B. Let $U = U_{y_1} \cap ...\cap U_{y_k}$ Then if $(x',y') \in U \times B$, we have $y' \in V_{y_i}$ for some $i$, and certainly $x' \in U_{y_i}$ Hence $(x',y') \in U_y \times V_y$,, which is contained in some $W$in $\mathcal{O}$
I am happy with the first paragraph of the proof. But it seems to me that the use of the compactness of $B$ in the second paragraph is not necessary. My alternative second paragraph goes like this:
Take $U = \cup_y U_y$. Then $U$ is open because each of $U_y$ is open. Each of $U_y$ is a subset of one of the elements of $\mathcal{O}$ ($W$). So $U$ is covered by the finite cover $\mathcal{O}$. $ B$ is also covered by the finite cover $\mathcal{O}$. Thus $U\times B$ is covered by a finite subcover $\mathcal{O}$
I think this is likely to be wrong and i am missing something. Any clues?
Edit: found the problem : it actually make little sense saying O covers B or U. They don’t live in the same dimensions.
Here is one example where taking the union would not work. We will find $\{x\}\times \mathcal{B} \subseteq \mathbb{R}^{2}$.
Consider $ x = 0$, $\mathcal{B}=\{0,1\}$, so \begin{gather*}\{x\}\times \mathcal{B}= \{0\}\times\{0,1\} = \{(0,0), (0,1)\} \end{gather*} Let
\begin{gather*}V_{0} = (-\frac{1}{2},1)\times (-1,\frac{1}{2}) \end{gather*} and \begin{gather*}V_{1} =(-1,\frac{1}{2})\times(\frac{1}{2},2) \end{gather*} we have $(0,0) \in V_{0}$ and and $(0,1) \in V_{1}$. Thus \begin{gather*}\mathcal{O} := \{V_{0},V_{1}\}\end{gather*} covers $\{x\}\times \mathcal{B}$.
Suppose you chose $U_{0}$ such that $ \frac{1}{2} \in U_{0}$. Then $ (\frac{1}{2},1) \in (U_{0}\cup U_{1}) \times \mathcal{B}$, but \begin{gather*} (\frac{1}{2},1) \not\in \Big( (-\frac{1}{2},1)\times (-1,\frac{1}{2})\cup (-1,\frac{1}{2})\times(\frac{1}{2},2) \Big) = V_{0} \cup V_{1} \end{gather*} So it is not possible to cover $(U_{0}\cup U_{1}) \times \mathcal{B}$ with any finite union of elements of $\mathcal{O}$.