Can this Lemma be extended a little? Consider this lemma (my question are below):

Lemma Given three pairwise orthogonal subspaces $X$, $Y$, $Z$ of a Hilbert space $H$ that span the whole space, any vector $\nu\in H,\ ||\nu||=1$, can be written as $$\nu=\alpha\nu_1+\beta\nu_2+\gamma\nu_3,\ \ \ \nu_1\in X,\ \nu_2\in Y, \ \nu_3\in Z,$$ with $\alpha,\beta,\gamma\geq0$ and $\alpha^2+\beta^2+\gamma^2=1$ and $\nu_1,\nu_2,\nu_3$ and unit vectors. 
Proof Let $\{x_j\}$, $\{y_j\}$, $\{z_j\}$ be orthonormal bases for $X$, $Y$, $Z$. Together, they form a basis for the whole $H$. So there exist coefficients such that
$$
\nu=\sum_j a_jx_j + \sum_jb_jy_j+\sum_jc_jz_j.
$$
As $\|\nu\|=1$, $\sum_j|a_j|^2+\sum_j|b_j|^2+\sum_j|c_j|^2=1$. Let
$$
\alpha=(\sum_j |a_j|^2)^{1/2},\ \beta=(\sum_j |b_j|^2)^{1/2},\ \gamma=(\sum_j |c_j|^2)^{1/2},\ 
$$
and
$$
\nu_1=\sum_j\frac{a_j}\alpha\,x_j,\ \nu_2=\sum_j\frac{b_j}\beta\,y_j,\ 
\nu_3=\sum_j\frac{c_j}\gamma\,z_j.
$$
Then $\nu_1,\nu_2\nu_3$ are unit vectors, $\alpha^2+\beta^2+\gamma^2=1$, and
$$
\nu=\alpha\nu_1+\beta\nu_2+\gamma\nu_3.
$$

My questions are:
1) Is this lemma true for complex Hilbert spaces ? (my guess would be "yes")
2) Is this lemma true, not for just three subspaces, but for subspace $X_1,\ldots X_n$, that are pairwise orthogonal and span the whole space ?
For who wants to know: This lemma is from  here.

The lemma holds for any finite number of real or complex Hilbert spaces of any finite or infinite dimension. The proof is unnecessarily complicated and involves some subtleties in the case of infinite-dimensional spaces that a straightforward proof doesn't need to get into; also it doesn't cover the case where one of $\alpha$, $\beta$, $\gamma$ is zero.
Since the subspaces $X_1,\dotsc,X_n$ span the space, there are vectors $x_i\in X_i$ such that
$$
\nu=x_1+\dotso+x_n\;.
$$
Take $u_i=x_i/\|x_i\|$ if $x_i\ne0$ and $u_i$ an arbitrary unit vector in $X_i$ otherwise. Then 
$$
\nu=\lambda_1u_1+\dotso+\lambda_nu_n
$$
with $\lambda_i=\|x_i\|$ if $x_i\ne0$ and $\lambda_i=0$ otherwise, so $\lambda_i\ge0$. Since the $u_i$ are orthonormal, $\|\nu\|=1$ implies $\lambda_1^2+\dotso+\lambda_n^2=1$.
