A question about the definition of a neighborhood in topology
I don't follow your argument; if anything, it strikes me as an argument for definition 2). You're saying that it may be non-trivial to show that a neighbourhood in the sense of 2) is a neighbourhood in the sense of 1). But why is the definition that contains more non-trivial properties the better one? It's easier to add the non-trivial properties when they're there than to subtract them out when they're not. If we use definition 2), we can easily express both concepts by speaking of "neighbourhoods" and "open neighbourhoods". If we use definition 1), we have to speak of "sets containing neighbourhoods" and "neighbourhoods" -- that's slightly more complicated.
Also, one often doesn't care whether the neighbourhood is open. For example, if a function is constant on a neighbourhood of $x\in\mathbb R$, its derivative at $x$ is zero. If you formulate this in terms of open neighbourhoods and for some reason you have that a function is constant on some closed interval, you have to insert the rather artificial step of specifying an open interval within that closed interval on which the function is constant. Or you could say that if a function is constant on a set containing a neighbourhood of $x\in\mathbb R$, its derivative at $x$ is zero, but that involves the complication with "sets containing neighbourhoods" again.
So, after all the great comments and answers (especially t.b.'s comment) I went and actually borrowed Bourbaki's General Topology from the library. After reading pages 18, 19, 20, I really have been able to get a better understanding of why definition 2) of a neighborhood is beneficial. As t.b., Andre and Joriki mention, it allows us to uniquely define a topology on a set X (proved as Proposition 2 in Bourbaki's book).
What I have been able to distill from all the comments and reading the first few pages of Bourbaki is that an important advantage of defining a neighborhood of a point $x \in X$ as a subset of $X$ that contains an open set containing $x$ is this simple consequence:
Any set that contains a neighborhood of $x$ is also a neighborhood of $x$. (This was mentioned by Joriki in his answer)
In fact note that if we define neighborhood as in 2) and denote by $B(x)$ the set of all neighborhoods of a point $x \in X$, then $B(x)$ satisfies the following properties:
(i) Any subset of $X$ that contains an element of $B(x)$ is an element of $B(x)$.
(ii) The intersection of two elements of $B(x)$ is an element of $B(x)$.
(iii) Every element of $B(x)$ contains $x$.
(iv) If $V \in B(x)$, then there exists $W \in B(x)$ such that for all $y \in W, V \in B(y)$.
Proposition 2 from Bourbaki now says the following: Let $X$ be a set, and for all $x \in X$ let $B(x)$ denote a collection of sets satisfying 1) - 4). Then there exists a unique topology on $X$ with respect to which the set $B(x)$ is precisely the collection of neighborhoods of $x$.
If one tries to prove this proposition, one will see that (i) is crucial in the proof of the fact that one can construct a unique topology just using neighborhoods (what t.b., Andre and Joriki have mentioned above), and it fails if we use definition 1) for a neighborhood. Note that conditions (ii), (iii), (iv) are satisfied by neighborhoods if defined in the sense of 1) (i.e., what people call open neighborhoods).
So, definition 2) puts open sets and neighborhoods on an equal footing. In the Introduction to General Topology, Bourbaki says that one can either take neighborhoods or open sets as the more primitive concept.
I also see why in a first course on topology, my professor did not decide to spend too much time explaining this issue.
(I do not mean to answer my own question this way. I am merely repeating what others have already said, as writing something out often helps me understand the concept better. So for those of you who have posted answers, please don't be offended.)
Let $X$ and $Y$ be two topological spaces. Denote by $\mathcal{V}_x$ the neighborhoods of $x \in X$. Do the same for $y \in Y$. Then, a function $f: X \to Y$ is continuous at a point $x \in X$ iff $$ f^{-1}\left(\mathcal{V}_{f(x)}\right) \subset \mathcal{V}_x. $$
The prototype for the concept of topological spaces is that of metric spaces. There, the natural concept is not that of "open set". In my opinion, the natural concept to make the link to the topology is that of a neighborhood base. The balls centered at $x$ are a neighborhood base at $x$. The same is true if you restrict yourself to balls of radius $\varepsilon_n > 0$, where $\varepsilon_n \rightarrow 0$.
With the concept of neighborhood, an open set is a set that is a neighborhood of all of its points. This is how one defines "open sets" in a metric space!! And this is only possible if the definition of "neighborhood" does not require that a set be open.
As you said, the fact that one has to worry about a set being or not open might be irrelevant, and yet, hard to prove. For example, for topological vector spaces or even topological groups, the topology is determined by the family of neighborhoods of a certain point. Linear transformations or homomorphisms are continuous iff they are continuous at a certain point. I see many professors struggling to prove that certain sets are open, when all they needed to do was to prove that they have non-empty interior!!