Metric spaces and openness

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.