About a special case of continuity in cartesian product of compacts result I want to prove the following statement:
Given $f: \mathbb{R^n} \rightarrow \mathbb{R}$, let $\phi:[0, +\infty) \rightarrow \mathbb{R}$ be  defined as $\phi(r) =sup_{|x| = r} f(x)$. Show that $\phi$ is continuous.
I was able to prove this result: "If $K$, $L$ are compact metric spaces and $f:K\times L \rightarrow \mathbb{R}$ is continuous, for each $y \in L$, put $\phi(y) = sup_{x\in K}f(x,y)$. Prove that $\phi:L\rightarrow \mathbb{R}$ as defined is continuous."
Can I generalize for $K_{1} \times K_{2} \times ...\times K_{n}$ and make it a special case of it (as in putting $\phi(r)$ as $\phi(x_{1},...,x_{n}) = supf(x_{1},..., x_{n})$)? (main question)
If you have any other (simple) ideas, please enlighten me.
 A: Since the topological product is associative (up to homeomorphism), the generalisation of your result to products with more than two factors is immediate once you know that the product of compact spaces is again compact (that even holds for products of infinitely many spaces).
But you look at the supremum (maximum) of $f$ on the sphere or radius $r$, and thus it doesn't help to look at $\mathbb{R}^n = \mathbb{R}\times \dotsc \times \mathbb{R}$. You want to use a product where one of the factors is a sphere. Thus let $S^{n-1} = \{ x \in \mathbb{R}^n : \lvert x\rvert = 1\}$ (whatever norm on $\mathbb{R}^n$ you use, the result is topologically the same), and look at the map
$$\mu \colon S^{n-1} \times [0,+\infty) \to \mathbb{R}^n,\quad (r,\xi) \mapsto r\cdot \xi.$$
Then $\mu$ is a continuous surjection (and the restriction to $S^{n-1} \times (0,+\infty)$ is a homeomorphism to $\mathbb{R}^n \setminus \{0\}$), and you can apply the result you proved to the map $\tilde{f} := f\circ \mu$ when you restrict to compact subsets of $[0,+\infty)$. Since continuity is a local property, you immediately obtain the continuity of $\phi$ from your result by noting that
$$\phi(r) = \sup \{ \tilde{f}(r,\xi) : \xi \in S^{n-1}\}.$$
