In metric space, show if a complement of a set is closed, then the set is open. I would appreciate if you could evaluate my sketch of the proof for the following problem:
For an arbitrary metric space $(X,d)$ and $Y\subset X$, show if $Y^c$ is closed, then $Y$ is open.
Sketch of the Proof:
(1) Given the metric space and $Y\subset X$, suppose $Y^c$ is closed.
(2) Suppose $Y$ is NOT open.
(3) By (2), $\exists y*\in Y$ s.t. $\forall\epsilon>0$ $B_\epsilon(y*)\not\subset Y$.
(4) By (3), $\forall\epsilon>0$ $\exists y_c\in Y^c$ s.t. $y_c\in B_\epsilon(y*)$.
(5) Since each open ball contains at least one member of $Y^c$, you can produce a sequence $(y_{c_n})\in Y^{c^\mathbb{N}}$ s.t. $y_{c_n}\rightarrow y*$.
(6) By (5), there exists a limit point $y^*$ that is not a member of $Y^c$.
(7) We assumed $Y^c$ to be closed so it contains all of its limit points.
(8) This is a contradiction.
(9) It follows that $Y$ must be open. QED.
 A: It's correct, but you can avoid contradiction and simplify the proof.
Suppose $Y^c$ is closed and let $y\in Y$. Since $y\notin Y^c$ and $Y^c$ is closed, there is a ball $B_r(y)$ such that $B_r(y)\cap Y^c=\emptyset$.
Therefore $B_r(y)\subset Y$.
Hence $Y$ contains a ball centered at each of its points.
A: I find it very helpful to understand the concepts and definitions to prove the following equivalent statement:
If $x $ is an interior point of $A $ then it is not a limit point of $A^c $ and if $y $ is a limit point of $A $ then it cannot be an interior point of $A^c $ and vice-versa.
The proof is direct from definitions.  If $x $ is interior to $A$ there is a neighborhood entirely in $A $ thus entirely disjoint from $A^c$ so it can not be a limit point of $A^c $.  Likewise if it's a limit point of $A $ all neighborhoods have points of $A $ so none are entirely within $A^c $ so it can not be an interior point of $A^c $.
Thus set $A$ is of the type that all points are interior points if and only if $A^c $ is of the type that $A^c $ contains all its limit points.  
Thus whether you define open as "consists of only interior points" and define closed as "its complement is open", or if you define closed as "contains its limit points", and define open as "its complement is open", or whether you define both directly, the definitions are equivalent.  i.e open $\iff$ consists only of interior points $\iff$ compliment contains all its limit points $\iff$ compliment is closed.
[It was late and I was redundant.  The strikeout text is repetitive and unnecessary.  However I choose to leave it in, as sometimes when dealing with new concepts and definitions the immediate contrapositive equivalences are always apparent.  (I managed to miss them last night.)]
