The statement is not true: $\mathbb{C}$ contains no discrete valuation ring having field of fractions $\mathbb{C}$, because a valuation of $\mathbb{C}$ must have a divisible value group. In particular this value group cannot be $\mathbb{Z}$.
The statement is true for example for every finitely generated extension field of $\mathbb{Q}$.
Sketch of the proof: let $L/K$ be a finite extension of fields, $v$ a valuation on $K$ and $w$ a prolongation of $v$ to $L$. Then $(w(L^\times ):v(K^\times ))\leq (L:K)$. In particular: if $v$ is discrete then $w$ is discrete.
Let $K/\mathbb{Q}$ be a finitely generated extension. Since all valuations on $\mathbb{Q}$ are discrete we are done if $K/\mathbb{Q}$ is algebraic.
If the extension is not algebraic it is a finite extension of a rational function field $\mathbb{Q}(T)$ in finitely many variables $T=\{t_1,\ldots ,t_n\}$. Thus it suffices to prove that the valuations of $\mathbb{Q}$ posses a discrete prolongation to $\mathbb{Q}(T)$. Such a prolongation is the Gauss prolongation of a valuation $v$. It assigns to a polynomial the minimum of the values of its coefficients.
Motivated by mr.bigproblem's answer I add the following:
A field $K$ contains a discrete valuation ring $O$ with field of fractions $K$ if and only if $K$ is the fraction field of a noetherian domain properly contained in $K$.
Sketch of the proof: the implication $\Rightarrow$ is obvious. If on the other hand $R$ is a noetherian domain, its integral closure $S$ in $K$ by the Mori-Nagata-theorem has the property that all localizations $S_p$ at primes of height $1$ are discrete valuation rings. Note that $S$ itself needs not be noetherian.
I think you will forgive me that I will not have a close look on all your arguments, but I can focus on the last statement from wikiepdia.
One direction is clear: If $(R, \pi)$ is a DVR, every fractional ideal is of the form $R\pi^{n}$ with $n \in \mathbb Z$, thus they are linearly ordered and in particular the intersection of two of those will always be the smaller one.
For the other direction, the property clearly implies that $R$ is local, because the intersection of two maximal ideals is a proper subset of both maximal ideals. More generally, the set of all fractional ideals is linearly ordered, because $I \cap J \in \{I,J\}$. In particular the set of all sub-vectorspaces of $\mathfrak m/\mathfrak m^2$ is linearly ordered, this of course enforces it to be one-dimensional. Hence $\mathfrak m$ is principal.
Summarizing, a DVR is a noetherian domain which is not a field and whose set of fractional ideals is linearly ordered. The statement on wikipedia with the intersections is just a reformulation.
Best Answer
Valuation rings have the property that for all elements $a,b$ we have $a|b$ or $b|a$. It follows easily that every finitely generated ideal is principal. In particular, a noetherian valuation ring has a principal maximal ideal. Besides, Krull's Theorem states that $\bigcap_{n \geq 0} \mathfrak{m}^n=0$ in a local noetherian ring $(R,\mathfrak{m})$. Therefore, 2 follows from 1.