Proof: $C(X×Y)=C(X)⊗C(Y)$ Where I can find the proof of the following theorem:
Let $X$ and $Y$ be compact Hausdorff spaces, $C(X)$ and $C(Y)$ the space of continuous functions on $X$ and $Y$ respectively, then we have $C(X×Y)=C(X)⊗C(Y)$.
 A: I assume that by $A \otimes B$ you mean the maximal tensor product of two C*-algebras $A,B$ (which is a completion of the tensor product of the underlying vector spaces). Let us restrict to the unital case. Then there are homomorphisms of C*-algebras $\iota_A : A \to A \otimes B$, $a \mapsto a \otimes 1$ and $\iota_B : B \to A \otimes B$, $b \mapsto 1 \otimes b$ which commute (i.e. $\iota_A(a)$ commutes with $\iota_B(b)$ for every $a \in A$, $b \in B$), and these are universal with respect to this property: If $f : A \to C$ and $g : B \to C$ are homomorphisms of unital $C^*$-algebras which commute, then there is a unique homomorphism $h : A \otimes B \to C$ of unital $C^*$-algebras such that $h \circ \iota_A = f$ and $h \circ {\iota_B} = g$.
It follows that if $A,B$ are commutative, then also $A \otimes B$ is commutative, and it is the coproduct of $A$ and $B$ in the category of commutative unital C*-algebras.
The category of compact Hausdorff spaces is equivalent to the dual of the category of commutative unital C*-algebras - via the functor given by $X \mapsto C(X)$ on objects and by $f \mapsto f^*$ on morphisms (commutative Gelfand-Naimark Theorem). Every equivalence of categories preserves products, and products in a dual category correspond to coproducts. This proves $C(X \times Y) \cong C(X) \otimes C(Y)$
A: Martin Brandenburg's answers gives a categorical proof using the full strength of Gelfand--Naimark.
One could also try to prove this more directly (using some of the ingredients that come into the proof of Gelfand--Naimark).
There is an obvious morphism $C(X)\otimes C(Y) \to C(X\times Y).$  (Here I just mean the usual tensor product.)  Now the Stone--Weierstrass approximation theorem
will show that this has dense image, and hence that the map from the completed tensor product $C(X) \widehat{\otimes} C(Y) \to C(X\times Y)$ is surjective.
[Added: actually, even this surjectivity isn't clear, as Martin Brandenburg
points out in a comment.]
To show that you get an isomorphism will take a little more work.

Another approach would be to show that the natural map $C(X) \widehat{\otimes} V \to C(X,V)$ (continuous $V$-valued functions) is an isomorphism for some class
of Banach spaces $V$, and to then apply this with $V = C(Y)$, and then use
that $C(X,C(Y)) = C(X\times Y).$  Off the top of my head, though, I'm not
sure how generally this formula holds.  (I don't see why it wouldn't hold for
arbitrary $V$, but I've not thought much about it, and my experience with this
kind of question comes from the $p$-adic context, where things are normally easier.)
