If $U$ is unitary, how can I show that there exist $w_{1},w_{2},...,w_{k}\in \mathbb{C}^{n}$, $k\leq n$, and $\theta_{1},\theta_{2},...,\theta_{n}\in \mathbb{R}$ such that
$U=U_{w_{1}}U_{w_{2}}\cdots U_{w_{k}} \begin{pmatrix} e^{i\theta_{1}} & 0 & \cdots & 0 \\ 0 & e^{i\theta_{2}} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & e^{i\theta_{n}} \end{pmatrix} $
where $U_{w_{i}}=I-\frac{2w_{i}w_{i}^{*}}{w_{i}^{*}w_{i}}$ are householder matrices.
a) Householder matrices are unitary; b) you can choose them to make the right-hand matrix upper-triangular; and c) a unitary upper-triangular matrix is diagonal. In more detail:
a) Householder matrices are unitary:
$$ \begin{align} \left(I-2\frac{ww^\dagger}{w^\dagger w}\right)\left(I-2\frac{ww^\dagger}{w^\dagger w}\right)^\dagger & =\left(I-2\frac{ww^\dagger}{w^\dagger w}\right)\left(I-2\frac{ww^\dagger}{w^\dagger w}\right) \\ & =I-4\frac{ww^\dagger}{w^\dagger w}+4\frac{ww^\dagger ww^\dagger}{w^\dagger ww^\dagger w} \\ & =I-4\frac{ww^\dagger}{w^\dagger w}+4\frac{ww^\dagger}{w^\dagger w} \\ & =I\;. \end{align} $$
b) You can choose them to make the right-hand matrix upper-triangular:
It suffices to show that you can produce zeros below the diagonal in the first column; the remaining columns can be dealt with in the same way, with the identity in the rows already fixed to preserve the zeros there. So let $v$ be the first column of $U$, and choose $w=v+\lambda e_1$, where $e_1$ is the unit column vector with a $1$ in the first row. Then
$$ \begin{align} \left(I-2\frac{ww^\dagger}{w^\dagger w}\right)v & =\left(I-2\frac{(v+\lambda e_1)(v+\lambda e_1)^\dagger}{(v+\lambda e_1)^\dagger(v+\lambda e_1)}\right)v \\ &=v-2\frac{v^\dagger v+\lambda v_1}{v^\dagger v+2\lambda v_1+\lambda^2}(v+\lambda e_1)\;. \end{align} $$
The fraction is $1$ for $\lambda=0$ and $0$ for $\lambda\to\infty$. Thus it takes the value $\frac12$ for some value of $\lambda$, and for that value, the resulting column vector is proportional to $e_1$, that is, it has zeros below the diagonal.
c) A unitary upper-triangular matrix is diagonal:
This follows directly from the fact that the scalar products of columns of a unitary matrix vanish.
The matrix obtained by left-multiplying $U$ by Householder matrices is upper-triangular; it is a product of unitary matrices, and thus itself unitary; thus it is a diagonal unitary matrix; and such a matrix must have the form of the right-most factor in the question.