Question: I came upon the following question (probably trivial...): 

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> 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 $X$?
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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.

Answer: 
Here is another ways of visualizing the problem. First, we arrange $16$ blue balls and $10$ green balls so that no two of the green balls are consecutive. Then we label the balls. 

Line up $16$ blue balls, leaving spaces between successive balls and at the ends of the row. There are $17$ such spaces, $15$ between successive blue balls and two at the ends of the row.
$$\square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square$$
To ensure that no two of the ten green balls are adjacent, we choose $10$ of these $17$ spaces in which to insert a green ball, which can be done in 
$$\binom{17}{10}$$ 
ways. Next, we label the balls from left to right. The letters on the green balls are the desired subset of the alphabet in which no two of the selected letters are consecutive. For instance, 
$$\color{green}{\bullet} \color{blue}{\bullet} \color{green}{\bullet} \color{blue}{\bullet} \color{blue}{\bullet} \color{blue}{\bullet} \color{green}{\bullet} \color{blue}{\bullet} \color{blue}{\bullet} \color{green}{\bullet} \color{blue}{\bullet} \color{green}{\bullet} \color{blue}{\bullet} \color{green}{\bullet} \color{blue}{\bullet} \color{blue}{\bullet} \color{blue}{\bullet} \color{green}{\bullet} \color{blue}{\bullet} \color{green}{\bullet} \color{blue}{\bullet} \color{green}{\bullet} \color{blue}{\bullet} \color{green}{\bullet} \color{blue}{\bullet} \color{blue}{\bullet}$$
corresponds to the selection $\{A, C, G, J, L, N, R, T, V, X\}$.
Having chosen 10 non consecutive letters from the alphabet, see how the gaps created by them will look like. There will be 10 gaps such that no two are adjacent and 16 letters. Thus the number of ways of selecting 10 non consecutive letters is same as putting 10 "gaps" in a sequence of 16 letters so that no two are consecutive. Thus we must choose from the 15 in-between gaps among the 16 letters and the two at the ends. Thus the number of ways is $\binom{17}{10}$