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Based on what I have understood by the definition of an accumulation point:

(N* is a deleted delta neighbourhood of M)

M is an accumulation point of S, a subset of R, if for all N*(M; delta) there exists an m in S such that m is in N*(M; delta).

It would seem that every point within S would satisfy this condition. Am I not seeing something?

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EDIT: Definition of a deleted delta neighbourhood (as requested):

For x in R and delta > 0, N*(x; delta) = {y in R: 0<|x-y|< delta}

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    Can you add the definition of a deleted delta neighborhood and an accumulation point that you are using? Borg if these have multiple usages. – 2017-02-20
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    I added the definition of deleted delta neighbourhood. The definition of an accumulation point is already included in the description. – 2017-02-20
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    What does $N*(M,\delta)$ mean? $M$ is a set, not a point right? I think ou didn't quite finish typing out the definition of $N*$ – 2017-02-20
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    Sorry, the formatting messed it up. Everything is there now. – 2017-02-20
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    "M is an accumulation **point** of S, a subset of R, if for all N*(M; delta) there exists an m in S such that m is in N*(M; delta)" – 2017-02-20

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So, if $S=\mathbb{R}$ (or any interval or union of intervals) then yes, every point is an accumulation point (more commonly called a limit point). However, when $S$ has isolated points, for example $S=\cup\{\frac{1}{n}:n\in\mathbb{N}\}$, this no longer holds. Consider $1\in S$ and $\delta=\frac{1}{3}$. The deleted delta neighborhood around $1$ of radius $\frac{1}{3}$ certainly contains many real numbers, but none of them are in $S$.