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I've tried this using open balls, but i get stuck.

Let $x\in(A')'$ then for all $r>0,\; B_r(x)-\{x\}\cap A'\ne\emptyset,\;$ so there exists $y\ne x$ such that $\|x-y\|0$ $B_s(y)-\{y\}\cap A\ne 0$.

I've been thinking I should use proof by contradiction by supposing that $x\notin A'$ but I don't know how to proceed. Some suggestion or hint? I can't use the closure and neither the frontier.

Note: $A$ is any set and $A'$ is the set of limit points of $A$.

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    I answered a question very similar to this http://math.stackexchange.com/questions/2141845/a-question-about-an-exercise-in-the-basic-topology-chapter-in-baby-rudin/2141874?noredirect=1#comment4405883_2141874. It was put on hold though – 2017-02-14
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    It might be wise to explain notation. I assume by $A'$ you are denoting the limit points of $A$? If that's not correct, say so, and even if it is correct, it wouldn't hurt to edit your post to say something like: $A'$ denotes: ________ – 2017-02-14
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    Yeah, I first assume A' meant the interior of A.... – 2017-02-14
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    Take open balls but smaller radii. For all $r$ there is $r/2$ and $B_{r/2}(x) - \{x\}$ has a point y $\ne x$ so that $y$ a limit point of A. Then take $s < d(x,y)$ and $B_s(y)$ has a point $w \in A$ but $d(y,w)< s = d(y,x)$ so $w$ isn't $x$ and $d(w,x) < d(w,y)+d(w,y) < s + r/2 < r/2 + r/2 = r$. So $x$ was a limit point. – 2017-02-14
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    A is any set and A' is the limit points set – 2017-02-14
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    Let $x\in (A')'$. We wish to prove that every neighborhood of $x$ has a member of $A$ in it different from $x$, so that $x\in A'$. Such a neighborhood $U$ has a member $y$ of $A'$ in it, different from $x$, because $x\in (A')'$. $U$ is a neighborhood of $y$ too, so it contains a member $z$ of $A$. This is almost what we want, except we might have $z=x$. To avoid $z=x$, note that $U\setminus \{x\}$ is also a neighborhood of $y$, and there has to be a $z\in A$ within $U\setminus \{x\}$. Balls just complicate things. – 2017-02-14
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    A contradiction is possible but not recommended. If There is a limit point x of (A')' that is not a limit point of A. Then there is a radius r so that B_r(x) has no point of A. Then B_r/2(x) has no points of A but must have points of A'. Let y be one of them. Then y is a limit point A. But B_r/2(y) $\subset$ B_r(x) and B_r(x) has not points of A. Os B_r/2(y) doesn't either. So y is not a limit point after all. – 2017-02-14
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    I've already editing muy post, but the WiFi conection is too bad, but don't worry, I accept your -1 this hints are too helpfully – 2017-02-14
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    No worries, Ragnar1204, such things (bad WiFi connections) do happen. I've edited your post; if it says what you mean, no need to edit it. – 2017-02-14

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Do what you intended but take increasingly smaller radii.

Let $x \in (A')'$ then for $r > r/2 > 0$, there is a $y \in A'$ so that $d(y,x) < r/2$. $y$ is a limit point of $A$ so for $s = d(x,y)< r/2$ there is a point $w \in A$ so that $d(y,w) < s < r/2$. So $d(x,w) \le d(x,y) + d(y,w) < r/2 + s < r/2 + r/2 = r$.

So $x$ is also a limit point of $A$.