Limit points of a subset. Let $X$ be a metric space, and let $E$ and $K$ be two sets such that $E\subset K$.
I want to prove:
If $p$ is a limit point of $E$, then $p$ is a limit point of $K$.
Proof: if every neighborhood of $p\in E$ contains a point $q\neq p$ such that $q\in E$, then this $q$ is also in $K$, then $p$ is a limit point of $K$. Is that correct?
 A: You have the right idea, but I think a piece is missing.
Yes, every neighborhood of $p$ in $E$ is a neighborhood of $p$ in $K$, but not every neighborhood of $p$ in $K$ is a neighborhood of $p$ in $E$. You have to start with a neighborhood of $p$ in $K$ and show that that neighborhood of $p$ in $K$, not just in $E$, contains a point of $K$ that is not $p$; you can say that a neighborhood of $p$ in $K$ contains a neighborhood of $p$ in $E$, but I think in this proof this assertion would need explanation.
A proof could start with 

Suppose that $B(p,\varepsilon_k) \subseteq K$ is a neighborhood of $p$ in $K$.

At this point you haven't explained why there needs to be an element of $E$ in that neighborhood; there may be subsets of $K$ that do not intersect with $E$, and you need to explain why this neighborhood is not one of them; to do this, you can look at the intersection of a neighborhood of $p$ in $K$ and a neighborhood of $p$ in $K$:
Let $B(p, \varepsilon_e) \subseteq E$ be a neighborhood of $p$ in $E$, and let $\varepsilon=\min\{\varepsilon_k, \varepsilon_e\}$. Then the intersection $B(p,\varepsilon_k) \cap B(p,\varepsilon_e)$ equals $B(p, \varepsilon)$, so it is a neighborhood of $p$ in both $E$ and $K$.
Since it is a neighborhood of $p$ in $E$ and $p$ is a limit point of $E$, it contains some element $q \in E$.
So we have $q \in B(p, \varepsilon) \subseteq K$, which means we have shown that an arbitrary open neighborhood of $p$ in $K$ contains another point of $K$ that is not $p$, which means $p$ is an accumulation point of $K$.
A: I dont think so. To proceed, let $p$ be a limit pointn of $E$. Then any neighbourhood of $N\setminus\{p\}\cap E\neq \emptyset.$  Now let $N$ be a neighbourhood of $\{p\}$ in $K$. Then $N\cap E$ is a neighbourhood of $p$ in $E$. So $(N\cap E\setminus \{p\})\cap E\neq \emptyset$ and therefore $N\cap K\neq \emptyset$. So $p$ is a limit point of $K$.    
