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I hope this is close to what you are asking. The following compactness principle turns out to be useful in certain construction in dynamical systems and in probability (in particular, in the theory of exchangeable random variables), and it may be seen as a topological version of the infinitary Ramsey theorem.

Lemma. Let $X,d$ a compact metric space, and $u:\mathbb{N}^2\to X$ a double sequence in $X$. Then, there exists a strictly increasing $\sigma:\mathbb{N}\to\mathbb{N}$ such that, denoting $v(i,j):=u(\sigma(i),\sigma(\, j)),$ the following limits exist, and coincide:

$$\lim_{i\to\infty} \lim_{j\to\infty} v(i,j) \ = \lim_{ i< j,\, (i,j)\to\infty} v(i,j).$$

The proof is just routine (iterated) application of the usual diagonal argument for sequences. How does it implies the infinitary Ramsey theorem? Take $X$ a discrete space of colors and $u$ a coloring an $X$-coloring of the complete graph with vertices set $\mathbb{N}$. Then the existence of the limit in the RHS means that the set $\{\sigma(i):i> c\}$ for some c is a monocromatic complete subgraph.

One can even state a more general version for multi-sequences $u(i)$ indicized on increasing $n$-ples $i:=(i_1< i_2\dots < i_n)$ of natural numbers; the game is that any parenthesization

$$(i_1,\dots,i_{k_1}),(i_{k_1+1},\dots,i_{k_2}),\dots,(i_{k_r},\dots,i_n)$$

produces a different iterated way of letting $i$ go to infinity (like when making the beads sliding from left to right in an abacus: in small clusters, one at a time, or all together ). The corresponding limits for u(i) may or may not exist and/or coincide; but, up to a selection of indices via a strictly increasing $\sigma:\mathbb{N}\to\mathbb{N}$, all these iterated limits do exist and coincide.

Actually, it may be argued whether it really gives a fundamentally different proof of Ramsey theorem as you are asking. Nevertheless, if you try submitting it to an analyst or to a geometer, I think you are much more likely to obtain an immediate proof of it than with the original set theoretic version (please confirm my guess). On the other hand, it may sound quite weird to a pure set theorist (do not necessarily confirm this statement).

1

I hope this is close to what you are asking. The following compactness principle turns out to be useful in certain construction in dynamical systems and in probability (in particular, in the theory of exchangeable random variables), and it may be seen as a topological version of the infinitary Ramsey theorem.

Lemma. Let $X,d$ a compact metric space, and $u:\mathbb{N}^2\to X$ a double sequence in $X$. Then, there exists a strictly increasing $\sigma:\mathbb{N}\to\mathbb{N}$ such that, denoting $v(i,j):=u(\sigma(i),\sigma(\, j)),$ the following limits exist, and coincide:

$$\lim_{i\to\infty} \lim_{j\to\infty} v(i,j) \ = \lim_{ i< j,\, (i,j)\to\infty} v(i,j).$$

The proof is just routine (iterated) application of the usual diagonal argument for sequences. How does it implies the infinitary Ramsey theorem? Take $X$ a discrete space of colors and $u$ a coloring of the complete graph with vertices set $\mathbb{N}$. Then the existence of the limit in the RHS means that the set $\{\sigma(i):i> c\}$ for some c is a monocromatic complete subgraph.

One can even state a more general version for multi-sequences $u(i)$ indicized on increasing $n$-ples $i:=(i_1< i_2\dots < i_n)$ of natural numbers; the game is that any parenthesization

$$(i_1,\dots,i_{k_1}),(i_{k_1+1},\dots,i_{k_2}),\dots,(i_{k_r},\dots,i_n)$$

produces a different iterated way of letting $i$ go to infinity (like when making the beads sliding from left to right in an abacus: in small clusters, one at a time, or all together ). The corresponding limits for u(i) may or may not exist and/or coincide; but, up to a selection of indices via a strictly increasing $\sigma:\mathbb{N}\to\mathbb{N}$, all these iterated limits do exist and coincide.

Actually, it may be argued whether it really gives a fundamentally different proof of Ramsey theorem as you are asking. Nevertheless, if you try submitting it to an analyst or to a geometer, I think you are much more likely to obtain an immediate proof of it than with the original set theoretic version (please confirm my guess). On the other hand, it may sound quite weird to a pure set theorist (do not necessarily confirm this statement).