Please check if my proof is right or wrong. Any advice is welcome. Moreover how can I check my proofs right or wrong when I solve some problem by myself?
(1) $P$ is closed
Let's assume point $z$ is a limit point of $P$ but $z$ is not a point of $P$. Since $z$ is a limit point, for all $r$ $p_i \in N_r(z)$ satisfies where some points $p_i\in P$. Let $\mathcal{E} (2) every point of $P$ is a limit point Assume every point of $P$ is not a limit point i.e. there exists r that $N_r(p_i)$ doesn't contain points in $P$ (other than $p_i$). By definition of condensation set $N_r(p_i)$ contain uncountably many point of $E$. By assumption, all points in $N_r(p_i)$ that neighborhood of that points contain countably many point of $E$ (if there exists a point that neighborhood is uncountably many point of $E$, this point should be in set $P$). Then $N_r(p_i) \subset \bigcup_x^\inf N_{r'}(x)$ Union of countable set is countable so $N_r(p_i)$ should be countable which contradicts to definition of condensation set. by (1) and (2) $P$ is perfect. (3) $P^c \cap E$ is at most countable. Following hint and previous exercises of pma 2.22,2.23, $R^k$ is seperable (2.22) and every seperable metric space has a countable base (2.23). So we can think ${V_n}$ as hint. $P$ is nonempty perfect set in $R^k$ so $P$ is uncountable (pma theorem 2.43). It is obvious that $P \subset W^c$ (if $x \in P$ then $x \in W$, it contradicts that $P$ contains uncountably many points of $E$). Conversly suppose $x \in W^c$. Then x is a point of $V_n$ where $E \cap V_n$ is uncountable. Since $V_n$ is a base and for any neighborhood $N(x)$ of $x$, $x \in V_n \subset N(x)$. Thus $E \cap N(x)$ is also uncountable that means $x \in P$. So $W^c \subset P$. Thus $P=W^c$