Metric spaces and openness I am asked to prove two things. I would like to know if the proof was elaborate and concise.
I would also like to know if proving reductio ad absurdum is looked down upon. I have heard from my professor say it's not recommended, but my analysis book says it is an invaluable tool.
a) Prove that in a metric space the complement of a point is open.
Suppose that the point $x$ is contained in a closed interval [x]. Now suppose for contradiction the complement of the point is closed, that is, $a<=x$ and $x<= b$. Now, the absurdity is at hand: $[x] \cup [a, b] \leftrightarrow [a, b]$ when they were originally meant to be disjoint. So the complement of the point must be open, that is, $a < x < b$.
b) Prove that any set in a metric space is an intersection of open sets. 
No clue here. I would gladly appreciate any clues. I am self-learning this.
 A: Let $(X,d)$ be a metric space.
Let $x \in X$.
We wish to show that $X\setminus \{x\}$ is open.
By definition, this means showing that for each $y \in X \setminus \{x\}$ there exists an $\epsilon\in(0,\infty)$ such that ….
Fix a specific $y$. Then the distance between $x$ and $y$ is $d(x,y)$.
Let $\epsilon = \text{something or other to do with }d(x,y)$.
Then …
A: You can do these exercises without reducing to contradiction. They are intended to ensure you understand the concepts and can apply definitions. I would guess that's perhaps why your professor discouraged use of proof by contradiction.
So here are some hints. Let your metric space be $X$. Note that since the context is a general metric space, we cannot appeal to any properties of $\mathbb{R}$, such as intervals (which require, among other things, an ordering). Instead, all we can work with is the distance function and things we've defined from the distance function, like convergent sequences and distance balls.
For part (a), you have a range of definitions of "open" and "closed" at your disposal. To prove that the complement in $X$ of a singleton $\{x\}$ is open, you might prove that for any $y\in X-\{x\}$, there is a small ball of radius $\epsilon > 0$ such that $x\notin B_\epsilon(y)$. Alternately, you might prove that $\{x\}$ is closed (does it contain every limit point?), and argue that the complement of a closed set is open.
For part (b), we are given an arbitrary set $A$. What sets could we intersect? As dfeuer suggests -- is to consider the complement of every point in the complement of $A$. We know these are open sets. Can we recover $A$ from them?
Again, the point here is to help you become comfortable with working in a more abstract setting. You have certain tools at your disposal, like the distance function and its derivative notions, while you cannot appeal to other, more familiar ideas, like intervals.

NB: In an earlier version of my answer, I had included the idea of "shrinkwrapping" $A$. As defeuer, Stefan H., and Brian M. Scott point out in the comments, that only produces the closure of $A$: any point in the boundary of $A$ will be contained in any $\epsilon$-neighborhood of $A$, hence will be in the intersection of all $\epsilon$-neighborhoods of $A$.
A: a) Prove that in a metric space the complement of a point is open.
Suppose there exists a metric space $x\in X$. We wish to show $X - {x}$ is open. Thus, define
$$\forall y \in M : D(x, y) < \epsilon$$
By definition, $ \forall y \in M, S_{\epsilon}(x)$ is an open ball.  Because open balls are open, it has been shown the complement of a point is open.
b) Prove that any set in a metric space is an intersection of open sets. 
Suppose there exists a set $E \subset M, E = E_{1} \cup E_{n}$. Next, suppose $e \in E$. Then $e \in E_{n} \forall n$. Since $E_{n}$ is open we have $B_{r_{i}}(e) \subset E_{n}$. If we take $r = min (r_1, r_2, ... r_n)$, then $B_{r}(e) \subset E$. QED.
