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Should we prove the fact "if K is compact in R^p, then K x {a} is also compact in R^(p+1)" ???

If so, Can We prove "if K is compact in R^p, then K x {a} is also compact in R^(p+1)" by using only definition of compactness ( in other words, by only using the fact that K is compact if every covering A of K can be replaced by a finite covering of K, using only sets in A )...

(I am studying compactness in analysis(book: Elements of Analysis - Robert G bartle), not topology... So It will be helpful not to use any topological knowledge if you can prove...)

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    The function $f:\Bbb R^p\to \Bbb R^{p+1} $ given by $f(x)=(x,a)$ is continuous and $f(K)=K\times \{a\}$,since $K$ is compact so is $K\times \{a\}$ – 2017-02-04

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Have you been taught that in $R^p$,compactness is the same as being closed and bounded? p.s. I didn't notice that you couldn't use Heine-Borel. In that case take any open cover $U_\alpha$ of $K\times(a)$. $U_\alpha\cap R^p$ will be an open cover of K.Choose a finite cover,and then at the most add one more open set to cover $K \times (a)$.

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    OP wants without that theorem – 2017-02-04
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    How does this work? you need to find open cover in $\Bbb R^{p+1}$ – 2017-02-04
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    Yes,those sets whose intersection with $R^p$ are finite open covers of K. – 2017-02-04
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If you take a sequence of the form $x_n\times a, x_n\in K$, you can find a converging subsequence $x_{n_k}$ of $x_n$ since $K$ is compact, $x_{n_k}\times a$ is a converging subsequence of $x_n\times a$.