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The statement is If $A subset mathbbR $ is any set, $I$ is an open set that contains a limit point of $L_A$ - the set of all limit points of $A$, then $I$ also contains a point of $A$. I approach this by letting $x$ in A with $x$ is a limit point of A. So for every $epsilon >0$ the deleted neighborhood $V^*(a) cap A$ is nonempty. From this how do I show that $x$ is also in $I$.

If $I$ is an open set containing a limit point of $A$, call it $x$ as you have. Then since $I$ is open it contains a ball around $x$, say $(x-epsilon,x+epsilon)$. Then since $x$ is a limit point of $A$ we know that as you've already pointed out $((x-epsilon,x+epsilon))cap Anot=emptyset$, so since $((x-epsilon,x+epsilon))subseteq I$ it follows that $((x-epsilon,x+epsilon))cap Asubseteq I$, hence $I$ contains an element of $A$.

Let $X$ be any topological space and $Asubset X.$ As per the comments by the proposer, a point $xin X$ is a limit point of $A$ iff every nbhd of $x$ contains a member of $A.$

Suppose $y$ is a limit point of the set $L_A$ of limit points of $A$. If $U$ is any nbhd of $y$ then there is an open $V$ with $yin Vsubset U.$

Now $V$ is a nbhd of $y$ so there exists $xin Vcap L_A.$ But $V$ is also a nbhd of $x$ (because $V$ is open and $xin V$ ), and $xin L_A$, so there exists $ain Acap V.$

So $ain Acap U$ because $Vsubset U.$

So any nbhd $U$ of $y$ contains a member of $A.$ So $yin L_A.$

This holds in every topological space.