Is it possible to write an arbitrary density matrix $\hat{\rho}$ in the following form ?
$$\hat{\rho} ~=~ \frac{1}{N} \sum_{\ell=1}^N \left|x_{\ell}\right\rangle \left\langle x_{\ell}\right|,$$
where $\left\{\left|x_{\ell}\right\rangle\right\}_{\ell = 1}^{N}$ are normalized states (but not necessarily orthogonal).
If yes, how can one prove this ?
Let us reformulate OP's question (v3) as follows:
Let $H$ be an $N$-dimensional Hilbert space. Is it possible to write an arbitrary density operator $$\tag{1} \hat{\rho}~\in~ B(H)~\cong~ {\rm Mat}_{N\times N}(\mathbb{C})$$ on the form $$\tag{2} \hat{\rho} ~=~ \frac{1}{N} \sum_{m=1}^N |m) (m|,$$ where $\left\{|m) \right\}_{m = 1}^{N}$ are normalized states $$\tag{3}(m|m) ~=~1, $$ but not necessarily orthogonal?
The answer is Yes.
Proof: Because $\hat{\rho}$ is a positive operator, it may be diagonalized wrt. an orthonormal basis. Hence there exists an orthonormal basis $\left\{|n\rangle \right\}_{n = 1}^{N}$, and eigenvalues $\lambda_1, \ldots, \lambda_N \geq 0$, such that
$$\tag{4} \hat{\rho} ~=~ \sum_{n=1}^N \lambda_n|n\rangle \langle n|,$$
and with unit trace
$$\tag{5} \sum_{n=1}^N \lambda_n~=~ {\rm tr} \hat{\rho}~=~1. $$
Now define
$$\tag{6} |m)~:=~ \sum_{n=1}^N \exp\left(\frac{2\pi i}{N} mn \right) \sqrt{\lambda_n} |n\rangle .$$
It is straightforward to check that eqs. (2) and (3) are satisfied.
My answer: not every density matrix is allowed to be written in this way. However, if you admit non normalized states, in finite dimensional Hilbert spaces, each density matrix can be expressed in this way and, moreover, there is an infinite number of decompositions with that form. (See for example the book Nielsen & Chuang "Quantum Computation and Quantum Information" )
