Subset of discrete and closed subset is also discrete and closed? I came upon the following question (probably trivial...): 

Let $X$ be a topological space Suppose $N \subseteq M \subseteq X$ and $M$ is closed and discrete in $X$. Then is it true that $N$ is closed and discrete in $X$?


My thoughts:  I believe the statement is true. 
Let $a \in N \subseteq M$, Then exists nhood $U_a \cap N = U_a \cap M = \{ a \}$. Thus $N$ is discrete. Suppose $N$ has an accumulation point; exists $x \in X$ such that $\{ x_n \} \subseteq N \subseteq M$ with $x_n \rightarrow x$. Then $x$ is also an accumulation point of $M$. As $M$ is closed, $x \in M$, contradicting discreteness. Hence, $N$ has no accumulation point, and is therefore closed. 
 A: You need to add that $\{ x_n \}$ is not eventually constant: 
[...] exists $x \in X$ such that $\{ x_n \}$ not eventually constant, $\{ x_n \}\subseteq N \subseteq M$ with $x_n \rightarrow x$. 
Otherwise you cannot conclude that $x$ is also an accumulation point of $M$, but only that $x$ is a limit point of $M$ (that is, isolated points are included).
Moreover, note that $\{ x_n \}$ could not be a sequence, but it is certainly a net, because you are talking here about topological space in general and not for instance about a second-countable topological space where sequence convergence is sufficient to caracterize set closure. So it would be better to write it as $\{ x_\alpha \}$, $\alpha \in A$, where $A$ is a directed set
A: Supposed N is not closed.  Then some x not in N with for all open U nhood x,
U cap N is not empty.  Since M is closed, x in M, which contradicts that
M is discrete.
A: The part that shows that $N$ is discrete is OK, but the second part is a bit confused to the end.
There's actually no contradiction in $x\in M$ (actually that follows from the fact that $M$ is closed and $x$ is a limit ponit of $M$). What you instead use is the fact that if $x_j\to x$ in a discrete space you must have an $N$ so that $x_j=x$ for every $j>N$. This means that if $x$ is a limit point of $N$ you have that $x_j=x$ (since it converges in $M$) eventually and since $x_j\in N$ we have that $N$ contains the limit point (and that $N$ contains all of it's limit points is the same as $N$ is closed).
To see that $x_j=x$ always eventually follows from the definition since there exists a nhood of $x$ that contains only $x$ and eventually $x_j$ must remain in that nhood.
