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For the (Hausdorff) compact spaces I can think of, compactness is established either using a product of compact spaces (including the Heine-Borel Theorem, the Banach-Alaoglu Theorem, Stone-Čech compactification, etc.) or by inheriting compactness from another space (e.g. the Hausdorff metric on compact subsets of another compact space). I guess it might be vague as to whether one-point/end type compactifications fall into the second category, but compactness is generally established from compact subsets of the original space.

Are there any examples of compact spaces whose compactness can be established via a fundamentally different method?

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    $\begingroup$ @YCor: Though on the other hand, $[0,1]$ can be realized as a continuous image of Cantor space $\{0,1\}^\omega$ (as can every compact metric space). $\endgroup$
    – Nate Eldredge
    Commented May 19, 2020 at 15:56
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    $\begingroup$ @YCor The usual proof of the Heine-Borel Theorem boils down to finding an infinite path in an infinite 0-1 tree. This is equivalent (in the reverse mathematics sense, over a weak subsystem of second-order arithmetic) to the compactness of $\{0, 1\}^\omega$. With this in mind, you can give an easier-to-remember proof of constructing the continuous function from $\{0, 1\}^\omega$ onto $[0, 1]$ by hand using binary expansion. $\endgroup$
    – Cameron Zwarich
    Commented May 19, 2020 at 16:01
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    $\begingroup$ If one allows continuous images, I guess every compact Hausdorff space is continuous image of a Stone space isn't it? So literally all that remain would be compactness of $\emptyset$, the singleton and the 2-point space...The question anyway seems to allow "different proof", not that it can't be proved in a given way. And of course "different proof" is not very well-defined (for sure, like most students when I was first taught compactness of $[0,1]$, I had never heard of compactness of $\{0,1\}^\omega$). $\endgroup$
    – YCor
    Commented May 19, 2020 at 16:09
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    $\begingroup$ Every compact Hausdorff space is homeomorphic to a closed subset of a hypercube. $\endgroup$
    – Michael Greinecker
    Commented May 19, 2020 at 17:25
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    $\begingroup$ For ordered topological spaces (like $[0,1]$ or $\omega_1 + 1$), compactness is just "extreme completeness" : every subset has a supremum. (including $\sup(X)=\max(X)$ and $\sup(\emptyset)=\min(X)$). $\endgroup$
    – Henno Brandsma
    Commented May 20, 2020 at 10:20

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For me the most basic compact spaces are $\{x_n: n \in \Bbb N\} \cup \{x\}$ when $x_n \to x$, (the countable cofinite space is a special case), of course all finite spaces, and all ordered topological spaces with a suprema for all subsets ("very order complete", usual order completeness being: "every nonempty subset that is bounded above has a supremum"). (finite sets can again be seen as a special case).

For all of these the compactness is a very basic fact, not derivable from a "product theorem".

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Elaborating on Michael Greinecker's comment: if $X$ is a compact Hausdorff space then the map $i \colon X \to \Pi_{C(X,I)} I$ given by $i(x)_f = f(x)$, where $I = [0,1]$, is a homeomorphism onto its image. So every compact Hausdorff space can be expressed as the closed subspace of the product of closed unit intervals, which is a partial negative answer to your question - establishing the compact Hausdorffness of a space by taking products and subspaces will, in a sense, always work.

That said, depending on your taste you might consider the Arzela-Ascoli theorem as an alternative source of compact Hausdorff spaces - this shows, for instance, that a uniformly bounded sequence of functions on Euclidean space with uniformly bounded derivatives forms a compact set. It's not so obvious how to see such a sequence as a closed subset of a product space, though this does become clearer when you follow through the proof of Arzel-Ascoli (as above, it sort of has to eventually).

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