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Most textbooks, (eg. Rudin's Principles of Mathematical Analysis) after defining open and closed sets, as well as the closure of a set, they move on to the theorem:

Let $B$ be a set in a metric space and let $\bar{B}$ be the closure of $B$. $B\cup\bar{B}$ is closed.

My question is: why is this a theorem, ie. that needs proof. Is it not obvious that if you add all limit points (the closure) to a set, the result is a set that contains all its limit points - ie. a closed set?

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    It is obvious to you, but may not be obvious to others. You understand that the limit points of $B$ are being added to it, others may not be able to look at it that way in the first glance. By the way, if this question is trivial for you, then that's good, because it should be after basic exposure to closed sets. But it certainly is a theorem, a non-trivial fact that needs proving. – 2017-01-17
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    One part of math is to be able to confirm your intuition with proof. – 2017-01-18
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    That's not what Rudin proved. Ift $\overline{B} = B \cup B'$ where $B'$ equal the limit points of $B$ then you need a theorem that $\overline B$ is closed, because it might be possible that there is a limit point of $\overline B$ that is neither a point of $B$ or a limit point of $B$. – 2017-01-18
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    " Is it not obvious that if you add all limit points (the closure) to a set, the result is a set that contains all **its** limit points" What is **it**? The original set or the new set? It contains the limit points of the original set, of course, but the new set may have new limit points that were not included. – 2017-01-18

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Suppose there were a point $x \not \in B$ so that every neighborhood of $x$ contains a limit point of $B$ but there is a neighborhood of $x$ that does not contain any points of $B$. Then $x$ is a limit point of $\overline B$ but $x$ is not a limit point of $B$. So $x \not\in\overline B$. So $\overline B$ is not closed.

Of course, the above scenario is impossible.

But we need a theorem to prove it.

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To see that $B\cup\bar B$ is closed you need to know two things: (1) that $B\subseteq\bar B,$ whence $B\cup\bar B=\bar B;$ and (2) that $\bar B$ is closed. How you prove (1) and (2) depends on how you defined "closure" and "closed set," which will vary from one textbook to another.

If you define the closure $\bar B$ as the intersection of all closed sets containing $B,$ then $B\subseteq\bar B$ is obvious, and $\bar B$ is closed because an arbitrary intersection of closed sets is closed.

If you define topology by the Kuratowski closure axioms, then $B\subseteq\bar B$ is an axiom. A closed set is defined as a set which is equal to its closure, and $\bar B$ is closed because $\bar{\bar B}=\bar B$ is another axiom.