Consider this result:
Let $(X,d)$ be a metric space, $E\subset X$ and $p\in E$. Prove that if $\,\forall r>0$ the set $\{x\in E : x\not\in B_r(p)\}$ is finite, then $E$ is compact
I am not asking for advice on how to prove it, I just want to know if this result has some more general implications, or interesting corollaries / equivalences
Thanks in advance
I would say that the real value of this result is in terms of the examples it paves the way for. Once we understand this result, we're led naturally to a collection of compact sets which one might not obviously think of: the (closed, bounded, and) scattered sets. This helps flesh out our understanding of compact sets in general. The situation is already interesting in $\mathbb{R}$ with the usual metric, so let's live there for simplicity; in particular, in $\mathbb{R}$ we get a very nice "taxonomy" of compact sets into four groups, three of which you already know and one of which is given by the compact scattered sets.
First, let's go back a step to compactness in general. We know that the compact subsets of $\mathbb{R}$ are exactly the closed and bounded sets (although this is definitely takes work, and quite fun, to prove). Amongst these, besides the trivial examples of the finite sets the most obvious examples are the closed and bounded intervals.
A bit harder to think about are the compact sets which are infinite but nowhere dense. The standard example of such a set is the Cantor set, which - together with its variants like the fat Cantor sets - is an important (counter)example in mathematics, especially analysis.
But not every infinite nowhere dense compact set looks like the Cantor set. Apart from these we also have the scattered sets. Basically, these are the "really nowhere dense" sets: a set $A$ is scattered iff for every nonempty $B\subseteq A$ there is some $x\in B$ which is isolated in $B$ - that is, such that there is some open ball $U$ around $x$ with $U\cap B=\{x\}$. This looks at first glance like discreteness, but it's actually much weaker: in $\mathbb{R}$, there are closed and bounded (hence compact) scattered sets which are infinite, whereas every closed and bounded discrete subset of $\mathbb{R}$ is finite.
For example, consider $\{{1\over n}:n\in\mathbb{N}\}\cup\{0\}$. This is obviously closed and bounded, and hence compact. It is also easy to check that it is scattered. Note that it fits the pattern in the OP: all points are isolated except one, namely the point $0$.
But that's only the tip of the iceberg. For example, it's a good exercise to whip up an example of an $A\subseteq\mathbb{R}$ which is closed, bounded, scattered, and whose set of accumulation points itself has exactly one accumulation point. This is like the previous example but "one level higher up." And we can keep climbing higher and higher - see the notion of Cantor-Bendixson rank.
And in a sense this actually is the end of the story of compactness, at least within $\mathbb{R}$. In $\mathbb{R}$ every compact set either:
is finite, or
is infinite but scattered (and hence countable), or
contains a copy of the Cantor set but no nontrivial closed interval, or
contains a nontrivial closed interval.
Moreover, in a precise sense this taxonomy has no "gaps:"
The countably infinite compact sets are exactly the infinite scattered sets, and every uncountable closed set in $\mathbb{R}$ contains a copy of the Cantor set, so in terms of cardinality there is no gap between the first and second or second and third bulletpoints.
Finally, from the third to the fourth bulletpoints we go up in topological dimension. It turns out that any closed subset of $\mathbb{R}$ which doesn't have dimension $0$ contains a nontrivial closed interval, so in terms of dimension there's no gap between the third and fourth bulletpoint.
Of course finer analyses of the compact subsets of $\mathbb{R}$ are interesting and important, but the above is still quite intuitively helpful.