$\{x\} \times B \subset \mathbb{R}^{n+m}$ is compact if $B \subset \mathbb{R}^m$ is compact and $x \in \mathbb{R}^n$ This is from Spivak Calculus on manifolds page 14. 
I cannot use the fact that the cartesian product of two compact sets which are subsets of Euclidean space is compact because he uses what is written in the title to prove that fact.
It seems that I am only allowed to use the definition of compactness here, which is that a set is compact if every open cover of it has a finite subvoer.
(The book only deals with Euclidean space here, and Spivak defines an open set $U$ to be such that for every element $x \in U$, there is an open rectangle $A$ such that $x \in A \subset U$.)
As he says in the book, this must be very easy to see, but I just cannot figure out. 
Thanks in advance.
 A: Well, I think this comes directly from the definition. For any set of cover $\{U_k\}_{k\in\mathcal{K}}$, then we can project it to the later $\mathbb{R}^m$ sub-space, set: $$
V_k:=\mathbb{P}(U_k),\quad\mathtt{where}\quad\mathbb{P}:\mathbb{R}^{n+m}\mapsto \mathbb{R}^m
$$
which maintains a set of open set in $\mathbb{R}^m$ and it forms a cover of B.
Due to B is compact, there exists a finite sub-cover $V_{k_i}$. Then by the axiom of Choice, pick up the corresponding $U_{k_i}$ and it forms a finite sub-cover of $\{x\}\times B$.
I think this should work. :)
A: The function $f:B\to \mathbb R^{m+n}$ defined by $y\mapsto x\times y$ is continuous. Therefore, as $B$ is compact so is $f(B)=\{x\}\times B$.
If you want to use covers, but avoid projections, then note that a cover of $\{x\}\times B$, endowed with the subspace topology, is a collection $\mathscr A=\{V_i\cap (\{x\}\times B):i\in I\}$ where $V_i$ are open sets in $\mathbb R^{n+m}$. Each $V_i$ is a union of basis elements $(U_1\times\cdots\times U_n)\times U_{n+1}\times\cdots\times U_{n+m}$ so we may assume without loss of generality that our cover is a collection of these in the first place. i.e. $\mathscr A=(U_{1i}\times\cdots\times U_{ni})\times U_{(n+1)i}\times\cdots\times U_{(n+m)i})_i:i\in I\}.$  Now, then $B\subseteq \bigcup_i (U_{(n+1)i}\times\cdots\times U_{(n+m)i})_i$   and as $B$ is compact, it is covered by finitely many elements $\{U_{(n+1)i}\times\cdots\times U_{(n+m)i}\}^N_{i=1}$. Then, ${x}\times B$ is covered by  $\{U_{1i}\times\cdots\times U_{ni})\times U_{(n+1)i}\times\cdots\times U_{(n+m)i}\}^N_{i=1}$.
A: Let $x = (x_1, \cdots, x_n)$.
Let $X$ be a subset of $\mathbb{R}^{m+n}$.
Let $P(X) := \{(y_1, \cdots, y_m) \in \mathbb{R}^m | (x_1, \cdots, x_n, y_1, \cdots, y_m) \in X\}$.  
Then, it is easy to prove that the following properties hold:  
$P(\{x\} \times B) = B$.
$P(\bigcup_{\lambda \in \Lambda} U_\lambda) = \bigcup_{\lambda \in \Lambda} P(U_\lambda)$.
If $X \subset Y$, then $P(X) \subset P(Y)$.  

Let $U_\lambda$ be an open set of $R^{m+n}$. Then $P(U_\lambda)$ is an open set of $R^m$.  

Proof:  
Let $(y_1, \cdots, y_m) \in P(U_\lambda)$.
Then $(x_1, \cdots, x_n, y_1, \cdots, y_m) \in U_\lambda$.
Since $U_\lambda$ is open, there exist $a_1, a'_1, \cdots, a_n, a'_n, b_1, b'_1, \cdots, b_m, b'_m \in \mathbb{R}$ such that $(x_1, \cdots, x_n, y_1, \cdots, y_m) \in (a_1, a'_1) \times \cdots \times (a_n, a'_n) \times (b_1, b'_1) \times \cdots \times (b_m, b'_m) \subset U_\lambda$.  
$(y_1, \cdots, y_m) \in (b_1, b'_1) \times \cdots \times (b_m, b'_m) = P((a_1, a'_1) \times \cdots \times (a_n, a'_n) \times (b_1, b'_1) \times \cdots \times (b_m, b'_m)) \subset P(U_\lambda)$.  
$\therefore P(U_\lambda)$ is an open set of $\mathbb{R}^m$.  

We now prove that $\{x\} \times B$ is compact.  

Proof:
Let $\{x\} \times B \subset \bigcup_{\lambda \in \Lambda} U_\lambda$.
Then $B = P(\{x\} \times B) \subset P(\bigcup_{\lambda \in \Lambda} U_\lambda) = \bigcup_{\lambda \in \Lambda} P(U_\lambda)$.
Since $B$ is compact, $B \subset P(U_{\lambda_1}) \cup \cdots \cup P(U_{\lambda_k})$.
Let $(x_1, \cdots, x_n, y_1, \cdots, y_m) \in \{x\} \times B$.
Since $(y_1, \cdots, y_m) \in B$, $(y_1, \cdots, y_m) \in P(U_{\lambda_i})$ for some $i \in \{1, \cdots, k\}$.
So $(x_1, \cdots, x_n, y_1, \cdots, y_m) \in U_{\lambda_i}$.
$\therefore \{x\} \times B \subset U_{\lambda_1} \cup \cdots \cup U_{\lambda_k}$.
So $\{x\} \times B$ is compact.
