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I define the sequence $x_n = \cos (x_{n-1}), \forall n > 0$.

For which starting value of $x_0 \in \mathbb{R}$ does the sequence converge?

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For any $x_0$, you must have $x_n \in [-1,1]$, for all $n>0$. Since $|\sin x| \leq \delta < 1$ for all $x \in [-1,1]$, we have that $\cos$ is a contraction, hence it has a unique fixed point on $[-1,1]$, and the sequence $x_n$ converges to this fixed point.

copper.hat
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Wikipedia article Fixed-point iteration says:

The fixed-point iteration $x_{n+1}=\cos x_n$ converges to the unique fixed point of the function $f(x)=\cos x$ for any starting point $x_0$. This example does satisfy the hypotheses of the Banach fixed point theorem. Hence, the error after $n$ steps satisfies $|x_n-x_0| \leq { q^n \over 1-q } | x_1 - x_0 | = C q^n$ (where we can take $q = 0.85$, if we start from $x_0=1$.) When the error is less than a multiple of $q^n$ for some constant $q$, we say that we have linear convergence. The Banach fixed-point theorem allows one to obtain fixed-point iterations with linear convergence.

Martin Sleziak
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This is just Banach's fixed point theorem, and it is appliable since

$$\forall\,x_1\,,\,x_2\in\Bbb R\,\,\,,\,\,\,|\cos x_1-\cos x_2|\stackrel{\text{M.V.T. for}\,\cos x}=|\sin c|\,|x_1-x_2|\leq |x_1-x_2|$$

for some $\,c\in\,(x_1\,,\,x_2)\,$

DonAntonio
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    I think that to apply Banach fixed-point theorem you need $|\cos u-\cos v|\le q|u-v|$ where $q$ is a constant such that $q<1$, but this can be done for $u,v\in[-1,1]$. BTW about the choice of notation - in the question $x_n$ is used for the recursively defined sequence, you use $x_{1,2}$ for arbitrary real numbers; this might be confusing for some readers. – Martin Sleziak Sep 24 '12 at 19:28