How to prove that a flat spacetime admits Minkowski coordinates?

That is a non-trivial theorem of (semi)Riemannian geometry based on Frobenius Theorem: if the Riemann tensor is everywhere zero, then every point belongs to a local chart where the metric has the standard constant diagonal form.

ADDENDUM

The idea of the proof is the following. One looks for vector felds $X$ such that $$\nabla X =0\:.$$ In coordinates $x^1,\ldots, x^n$, this leads to a first-order equation for the components of $X$.

Next, looking at the found equation and using the condition $Riemann =0$ everywhere written in terms of connection coefficients, Frobenius theorem for first-order PDEs in $\mathbb R^n$ proves that, in a neighborhood of any point $p\in M$, there exist such $X$ satisfying $X(p) = Z_p$ where $Z_p\in T_pM$ is arbitrarily fixed.

So, one can construct $n= \dim(M)$ vector fields $X_{(k)}$, $k=1,\ldots,n$ in a neighborhood $U$ of $p$ such that $\nabla X_{(k)}=0$ and $X_{(k)}(p) = Z_{(k)p}$. Since the scalar product is preserved (from $\nabla X_{(k)}=0$ and the fact that the connection is metric), if $g_p(Z_{(k)p}, Z_{(h)p})= \eta_{kh}$, we have that $g(X_{(k)}, X_{(h)})= \eta_{hk}$ constantly on $U$.

Finally, one has to look for coordinates $y^a= y^a(x^1,\ldots,x^n)$ around $p$ such that $$X_{(k)}= \frac{\partial}{\partial y^k}\:.$$ Writing down this equation in coordinates $x^1,\ldots, x^n$, applying once again Frobenius theorem it is possible to prove that these local coordinates do exist around $p$. This way, shrinking further $U$ around $p$ we end up with a coordinate system $y^1,\ldots,y^n$ covering it where the metric is constant: $$g\left(\frac{\partial}{\partial y^k},\frac{\partial}{\partial y^h} \right)= \eta_{kh}\:.$$

In general, this procedure cannot produce a global chart where the metric is constant. There are trivial counterexamples starting from Minkowski space and assuming some identifications to produce a flat compact torus. This manifold is flat but cannot be covered by a global chart otherwise it would be diffeomorphic to $\mathbb R^n$ which is not compact.

I think my answer is consistent with Valter’s, but low-faluting.

Parallel transport preserves inner products of vectors, and the vanishing of the Riemann tensor guarantees the path independence of parallel transport within any local chart. You can start with an orthonormal tetrad and extend the local chart and Minkowski metric from an arbitrary origin. The elements of the metric tensor are defined as pairwise inner products of basis vectors in the tetrad.

To extend the metric globally, exploit the homeomorphism to/from ${{R}^{4}}$ to induce a (possibly curvilinear) coordinate system in which every tuple $(t,x,y,z)$ corresponds to a point of your flat space. A convenient path to reach any point is $(t,x,y,z)=(as,bs,cs,ds)$ with constant coefficients and parameter s running from 0 to 1. This path also happens to be unique, but existence is the key.

Nutshell proof of path-independence: The argument relies on the fact that any path from point A to B is homotopic to any other path in a Euclidean space. The “history” of the continuous deformation sweeps out a two-dimensional surface, which may be broken into plaquettes. The Riemann tensor describes the change in a vector resulting from transport around an infinitesimal plaquette. Integrating the little zeroes gets you a big zero.

Constructing Minkowski cooordinates: Once the orthonormal tetrad of basis vectors or 1-forms has been carried unambiguously to the entire space, we can construct the four coordinate functions by doing path integrals such as $\Delta t=\int{d\mathbf{x}\cdot \mathbf{u}}$, where u is the 1-form that represents the gradient of time, and likewise for x,y,z. The result is path-independent per Stokes’s theorem, because parallel transport guarantees that the covariant derivative and hence the curl of u is zero.

I didn't actually realise that I have already solved the 2nd 'problem' for the Riemannian case but the proof most certainly does not carry over to the general Lorentzian. However, I am fairly confident that if you have a Lorentzian manifold $\mathcal M$ that can be written as a 'product of time and space' (more rigorously as a foliation) , $\mathcal M = \mathbb R \times M$ for a Riemanninan manifold $M$ (which is usually the case in physical applications) then the proof that I will give for Riemannian manifolds should carry over to this special case of Lorentizian manifolds:

Our assumptions for the Riemanninan case will be the following:

• $M$ a smooth manifold with vanishing curvature $Riem =0$ i.e. flat and
• there exists a homeomorphism $f:M\to \mathbb R^{\dim M }$ i.e. topologically 'trivial' (so in particular $M$ is simply connected).
• I will further assume that $M$ is complete (i.e. every convergent series has a limit in $M$).

There is a characterisation of locally symmetric spaces via Riemann curvature tensor that is (Cartan-Ambrose-Hicks Theorem):

$M$ is locally symmetric if and only if $\nabla Riem =0$.

In our case this is trivially satisfied. Furthermore, every simply connected complete, locally symmetric space is globally symmetric. Hence, $M$ is a simply connected globally symmetric space.

Luckily there is a classification (Cartan classification) of simply connected globally symmetric spaces, which states:

Let $M$ be a simply connected globally symmetric space. Then $M= M_0\times M_+ \times M_-$, where $M_0$ is an Euclidean factor and $M_\pm$ is a symmetric space of compact ($+$) and non-compact ($-$) type resp. Furthermore, the sectional curvature of $M_\pm$ is larger than (resp. smaller than) or equal to but not identically equal to 0.

Since the curvature is everywhere zero, we must have $M_\pm = \emptyset$ because otherwise the sectional curvature of $M_\pm$ would be identically zero. Hence $M= M_0=\mathbb R^{\dim M} $ $\square$