Prove 2-dimensional Riemann manifold is conformally flat

[PhyPsy]

Homework Statement

Establish the theorem that any 2-dimensional Riemann manifold is conformally flat in the case of a metric of signature 0.
Hint: Use null curves as coordinate curves, that is, change to new coordinate curves
\lambda = \lambda(x⁰, x¹), \nu = \nu(x⁰, x¹)
satisfying
g^ab\lambda_,a\lambda_,b = g^ab\nu_,a\nu_,b = 0
and show that the line element reduces to the form
ds² = e^2\mud\lambdad\nu
and finally introduce new coordinates \frac{1}{2}(\lambda + \nu) and \frac{1}{2}(\lambda - \nu)

Homework Equations

Conformally flat metric: g_ab = \Omega²\eta_ab (\eta_ab is a flat metric)
ds² = g_abdx^adx^b

The Attempt at a Solution

It says in the hint that the metric has a signature of 0, so it must be flat in a specific set of coordinates, but there are 3 different coordinate systems the hint tells me to use in this problem:
(x⁰, x¹), (\lambda, \nu), and [\frac{1}{2}(\lambda + \nu), \frac{1}{2}(\lambda - \nu)]
so I don't know in which coordinate system I should make the assumption that the metric is flat.

I know that the line element equation in the (\lambda, \nu) coordinate system is:
ds² = g_{\lambda\lambda}d\lambda² + g_\lambda\nud\lambdad\nu + g_\nu\lambdad\nud\lambda + g_\nu\nud\nu²
and if I assume the metric is flat in this coordinate system, then that equation can be reduced to:
ds² = g_{\lambda\lambda}(d\lambda² - d\nu²) + 2g_\lambda\nud\lambdad\nu
where I used the symmetry of the metric. If I am to reduce this equation to
ds² = e^2\mud\lambdad\nu
then I need to show that
1. g_{\lambda\lambda}d\lambda² + g_\nu\nud\nu² = 0
2. 2g_\lambda\nu = e^2\mu.

I don't know how to do either. I have substituted d\lambda with (\partial\lambda/\partialx⁰)dx⁰ + (\partial\lambda/\partialx¹)dx¹ and tried all sorts of algebraic manipulations, but I have not been able to cancel out those 2 terms or figure out how 2g_\lambda\nu = e^2\mu. Could I get some help, please?

 

[PhyPsy]

I've managed to work through the steps in the hint, but it didn't lead me to the g_ab = \Omega²\eta_ab equation that I should be getting. The only way I see to eliminate the d\lambda² and d\nu² terms in the line element equation is to assume that d\lambda² = d\nu². Since the (\lambda, \nu) coordinate system is arbitrary, I should be allowed to make this assumption.
Then, I set \mu = \frac{1}{2}\int(\partial\lambda/\partialx⁰)²[(dx⁰ + dx¹)/(dx⁰ - dx¹)], and this allows me to define g_\lambda\nu + g_\nu\lambda = e^2\mu. I came up with the equation for \mu from the identity \frac{d}{dx}ln x = \frac{1}{x}.

I did the coordinate transformation to [\frac{1}{2}(\lambda + \nu), \frac{1}{2}(\lambda - \nu)] and came up with this:
g_{\lambda\lambda} = g_yz/2
g_\nu\nu = -g_yz/2
g_\lambda\nu = g_yy/2
where y = \frac{1}{2}(\lambda + \nu) and z = \frac{1}{2}(\lambda - \nu)

There are 2 problems with this. One, g_{\lambda\lambda} should be a function of g_yy, g_\lambda\nu should be a function of g_yz, etc. if I am to show that g_ab = \Omega²\eta_ab.
Two, even if the above were true, the metric in the (\lambda, \nu) coordinate system is not flat, because if it was, then g_\lambda\nu would be 0, not e^2\mu/2, and ds² would be 0. So all I would have proven is that \overline{g}_ab = \Omega²g_ab, not g_ab = \Omega²\eta_ab. Is there someone who can help?