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From section 29 of Munkres' topology. The bar on top of the 'B' indicates that it is the closure of B. The example above assumes that $\mathbb{R}^{\omega}$ is in the product topology with respect to the standard topology of $\mathbb{R}$.

My understanding: To derive a contradiction for the local compactness of $\mathbb{R}$, the proof uses the fact that a closed set of a compact space is compact. Furthermore, if X is a compact space in Y and Y is a compact space in Z, then X is a compact space in Z. If we show that X (in this case, the closure of B) is not compact in Z ($\mathbb{R}^{\omega}$), then we have proven that $\mathbb{R}^{\omega}$ is not locally compact. This part follows from the fact that $\mathbb{R}^{\omega}$ is not compact.

My logic gap: The proof that none of $\mathbb{R}^{\omega}$'s basis elements is contained in a compact subspace seems to rely on the assumption that, if B lies in compact subspace C, then the closure of B also lies in C. To me, this seems to be the only way the above logic can be applied to arrive at the subsequent contradiction, but I can't see any obvious way to prove this.

Am I thinking about this wrong? Any proofs or guiding hints would be much appreciated.

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    $\begingroup$ Compact subsets in euclidean space are closed, and the closure is the smallest closed set above the set. $\endgroup$ – Randall Jul 14 '18 at 3:22
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    $\begingroup$ $\mathbb{R}^{\omega}$ is not Euclidean. However, Compact subsets are closed in any Hausdorff space, and so Randall's comment "works". $\endgroup$ – David Hartley Jul 14 '18 at 7:17
  • $\begingroup$ @DavidHartley You’re right. I was arguing in a coordinate slot but that’s not the right way. $\endgroup$ – Randall Jul 14 '18 at 12:43
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Community wiki answer so the question can be marked as answered:

As noted in the comments, the closure of $B$ is the intersection of all closed sets containing $B$; since $\mathbb R^\omega$ is Hausdorff, the compact set $C$ is closed; hence the closure of $B$ is contained in it.

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