Can the neighborhood definition of a limit be used to prove a limit exists? It's easy to show that the following is equivalent to the usual $\epsilon$-$\delta$ definition:

Let $f:D\subseteq \Bbb R\to \Bbb R$ and suppose $c\in \Bbb R$ is an accumulation point of $D$.  Then $$\lim_{x\to c} f(x) = L$$ iff for every neighborhood $V$ of $L$, there exists a punctured neighborhood $U$ of $c$ such that $$x\in D\cap U \implies f(x) \in V.$$

But I'm having trouble proving even a simple limit exists with this definition.  I realize that some definitions are made more for proving theorems than doing practical problems, and this is probably an example of that.  But I find this topological definition pretty attractive and would thus like to know for sure whether or not this definition is useless at proving that limits of a given $f(x)$ exist.
Edit: The topology on $\Bbb R$ is the standard topology and on $D$ it is the subspace topology induced by the standard topology on $\Bbb R$.
If one can use this definition, a (relatively) simple example of its use would be much appreciated.
 A: The details for a proof of this nature will depend essentially on what you know about $f$. The goal is to show that, given any $V$ an open neighborhood of $L$, that you can find an open set, $U$ for the topology of $D$, such that $f(U-\{ c\})\subseteq V$. In order to actually carry this out, you will need to have meaningful ways of representing the open neighborhoods of $L\in\mathbb{R}$ and those of $c\in D\subseteq\mathbb{R}$. This is what the notion of a basis for a topology is meant to do. A basis for a topology is a collection of open sets from which every open set can be constructed via set-theoretic unions ($\cup$). Note that I didn't not specify that the unions needed to be finite, hence infinite unions, even uncountable unions may be needed (most generally, this can be notated as:
$\underset{i\in I}{\bigcup}U_i$
Where the set $I$ serves to index the open sets being unioned (the $U_i$'s).
For the standard topology on $\mathbb{R}$, the set of open intervals, $\{ (a,b)|a<b\}$, is a basis for the topology, hence every open set in the standard topology for $\mathbb{R}$ can be expressed in the form
$\underset{i\in I}{\bigcup}(a_i,b_i)$
The open sets for $D$ are merely the intersections of those for $\mathbb{R}$ with $D$.
$D\cap(\underset{i\in I}{\bigcup}(a_i,b_i))$
Because every open set can be formed from the basis sets by unions only, it suffices to prove that the definition of limit you gave is satisfied for basis elements containing $L$. Actually, you only need consider those open intervals that have $L$ in the middle, since every basis element containing $L$ also contains an open interval (which is another basis element) with $L$ as its midpoint.  (this is what proves the equivalence of the topological definition of limit with the usual $\epsilon$-$\delta$ definition of limit).
In practice, actually showing that the required open sets in the domain exist so that $f(D\cap U)\subseteq V$ will be roughly the same whether you use the $\epsilon$-$\delta$ definition or the topological one. But the value of the latter is that it generalizes to arbitrary topological spaces, rather than being specific to $\mathbb{R}$ (an enormous generalization, I might add). I find that the topological definition is also easier to picture, and thus can help intuition for whether or not a given point is a limit point or not (note: I say a limit point rather than the limit point because in some topological spaces limit points need not be unique. (!) A sufficient condition for uniqueness of limit points is the Hausdorff property; distinct points can be separated by disjoint open sets).
A: $\forall \epsilon >0, \exists \delta > 0$ such that $|x-c|<\delta \implies |f(x)-L|< \epsilon$
$d(f(x),L) < \epsilon$ defines a neighborhood  $V$
and $d(x,c) < \delta$ is a neighborhood $U$ around $c$
For a neighborhood of radius $\epsilon$ about $L$ there exists a $U$ such that $x\in U$ (and $f(x)$ is defined, i.e. $x\in D)  \implies f(x)\in V.$
To prove it the other way.
Any neighborhood $V$ about $L$ there exists an open ball $B_\epsilon = \{y: d(y,L)< \epsilon\}$ such that $B_\epsilon \subset V$
Let $U = \{x: |x-c|<\delta\}$
For all V there exists a $B_\epsilon$ and for all $B_\epsilon$ there exists a $U$ such that $x\in U\implies f(x) \in B_\epsilon$. 
$\forall \epsilon>0, \exists \delta>0$ such that $d(x,c)<\delta \implies d(f(x),L) < \epsilon$
