Problem 1:
Let $(X,d)$ be a metric space. Let $(x_n)_n$ be a sequence in $X$.
If $x_n\rightarrow x$ then either $x$ is a limit point of $\{$ $x_n$ $:$ $n\in \mathbb{N}$ $\}$ or $x=x_n$ for infinitely many $n\in \mathbb{N}$ .
My proof:
Suppose $x$ is not a limit point of $A=$ $\{$ $x_n$ $:$ $n\in \mathbb{N}$ $\}$. This means that there exists a radius $r>0$ such that $B(x,r)\cap$ $A\backslash$ $\{$ $x$ $\}$ $=$ $\varnothing$. Since $x_n\rightarrow x$ , there exists $N\in \mathbb{N}$ such that whenever $n\geq N$, we have that $x_n\notin A$ $\backslash$ $\{$ $x$ $\}$. But since $x_n\in A$ for each $n$, it follows that $x_n=x$ for infinitely many $n's.$
Problem 2:
Let $(X,d)$ be a metric space and suppose $A\subseteq X$ has not limit points. Then $(A,d|_A)$ is topologically equivalent to a discrete space.
Proof: Suppose $A$ has no limit points, so each point in $A$ is an isolated point. Let $U$ be an arbitrary subset of $A$. For $x\in U$ , there exists $r>0$ such that $B_A(x,r)=B(x,r)\cap U$ . Hence $B_A(x,r)\subseteq U$. So $U$ is open in $A$.
Are my proofs correct?
The first proof is correct. If $B(x,r) \cap (A\setminus \{x\}) = \emptyset$ but for all $n \ge N$, $x_n \in B(x,r)$ this can only happen if $x_n = x$ for those $n$. So the sequence is eventually constant (the limit).
The second too: as $x \in A$ is not a limit point of $A$ it is indeed isolated in $A$ and so $\{x\}$ is relatively open in $A$. As this holds for all $x \in A$, $A$ has the discrete topology as its subspace topology. Moreover $A$ is closed ($A$ is closed iff $A' \subseteq A$ and here $A'=\emptyset$). So alternatively we could have said: any $B \subseteq A$ obeys $B' \subseteq A'=\emptyset$ so $B$ is closed in $X$ (and as all subsets of $A$ are closed (hence open too) it has the discrete topology.