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Show that every accumulation point of a set that does not itself belong to the set must be a boundary point of the set.

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    why you cannot prove this yourself? no disparaging but this is elementary enough.2012-11-07

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Call the set $S$. Denote this accumulation point as $p \not\in S$. Every neighborhood around $p$ contains a point of $S$ (because it's an accumulation point of $S$) and a point not in $S$ (namely, itself).

Therefore, $p$ is in the boundary of $S$. QED.

Of course, your proof will look different depending on the version of 'boundary' you're using. I was using the third definition here.