$(e_{n})$ orthonormal basis, $(f_{n})$ orthonormal sequence such that $\sum\left\|e_{n}-f_{n}\right\|^{2}<\infty$ Then $(f_{n})$ is orthonormal basis. Let $H$ be a hilbert space, $(e_{n})$ a orthonormal basis of $H$, and, $(f_{n})$  a orthonormal sequence on $H$ such that
$$\sum_{n=1}^{\infty}\left\|e_{n}-f_{n}\right\|^{2}<\infty. \tag{I}$$
Show that $(f_{n})$ is also a orthonormal basis.
Remark: My idea was the following: 
First I show the following fact:

Fact 1:  Let $H$ be a Hilbert space, $(x_{n})_{n=1}^{\infty}$ a orthonormal basis of $H$, and let $(y_{n})_{n=1}^{\infty}$ be a sequence in $H$ such that 
  $$\sum_{n=1}^{\infty}\left\|x_{n}-y_{n}\right\|^{2}<1. \tag{II}$$
  Then, if $z\bot y_{n}$ for all $n\in\mathbb{N}$, then $z=0$.

Let $f\bot f_{n}$ for all $n\leq 1$, then by (I) there exists $m$ such that
$$\sum_{n=m+1}^{\infty}\left\|e_{n}-f_{n}\right\|^{2}<1. \tag{III}$$
Therefore, by Fact 1 we have that $\left\{e_{1},\ldots,e_{m},f_{m+1},f_{m+2},\ldots\right\}$ is total in $H$.
I need to show that $\left\{f,f_{1},\ldots,f_{m}\right\}$ is linearly dependent.
 A: $\def\span{\operatorname{span}}$
This solution uses ideas from  Demophilus and  DisintegratingByParts , who removed his/her nearly complete solution for some unknown reason.
Take $m \in \mathbb{N}$ such that $\sum_{n=m+1}^\infty \|e_n-f_n\|^2  = c <1$ and take $V = \overline{\text{span}\{e_{m+1},e_{m+2}, \ldots\}}$.     Let $W = \span\{e_1, \dots, e_m\} = V^\perp$.  Let  $V' =  
\overline{\text{span}\{f_{m+1},f_{m+2}, \ldots\}}$. 
There is an isometry $A'$ from $V$ to $V'$ determined by $e_j  \mapsto f_j$  for $j \ge m+1$.   Define $A : H = W \oplus V \to H$ by $A( w + v) = w + A'(v)$.   The range of $A$ is $W + V'$. 
Then
$$
(A - I)(w + v) = A'(v) - v = \sum_{ j = m+1}^\infty   \langle v, e_j\rangle  (f_j - e_j).
$$
$$
\begin{aligned}
|| (A - I) (w + v)||   &\le   \sum_{ j = m+1}^\infty  | \langle v, e_j\rangle| \ \  ||  f_j - e_j||  \\
& \le   \left( \sum_{ j = m+1}^\infty  | \langle v, e_j\rangle|^2\right ) ^{1/2}
 \left( \sum_{ j = m+1}^\infty ||  f_j - e_j  ||^2\right )^{1/2}  \\
 & = c^{1/2}  ||v|| \le  c^{1/2}  ||w + v||, 
\end{aligned}
$$
where the very last estimate uses that $w$ and $v$ are orthogonal, so $||v + w||^2 = ||v||^2 + ||w||^2$.
Thus  $|| A - I ||\le c^{1/2} < 1$,     and $A$ is invertible. (See the proof sketch below.)   In particular $A$ is surjective so $W+ V' = H$.  Check that $W \cap V' = (0)$, so $H = W \oplus V'$  (not an orthogonal direct sum). 
In particular $H/V'$ is $m$--dimensional, so $(V')^\perp$ is $m$ dimensional.  But $\{f_1,\ldots, f_m\}$ is an orthonormal subset of $(V')^\perp$ of cardinality $m$,  so $\{f_1,\ldots, f_m\}$  is an orthonormal basis of $(V')^\perp$.   Therefore $\{f_j : 1 \le j < \infty\}$ is an orthonormal basis of $H$.
Note:  In any Banach algebra with identity $I$ (and $||I|| = 1$)  if $x$ is an element satisfying $||x|| < 1$, then $I - x$ is invertible with inverse $\sum_{j = 0}^\infty x^j$.  Apply this in our context, noting $A = I - (I - A)$.
A: Take $m \in \mathbb{N}$ such that $\sum_{n=m+1}^\infty \|e_n-f_n\|^2 <1$ and take $V = \overline{\text{span}\{e_{m+1},e_{m+2}, \ldots\}}$. By your fact 1 we have that $V = \overline{\text{span}\{f_m,f_{m+1}, \ldots\}}$. Notice that $V^\perp = \text{span}\{e_1,\ldots, e_m\}$. But we also have that $\{f_1,\ldots, f_m\} \subset V^\perp$ is a linearly independent subset. Because $\dim V^\perp =m$, $\{f_1,\ldots, f_m\}$ is a basis of $V^\perp$. Then we must have that $(f_n)_n$ is an orthonormal basis of $V \oplus V^\perp =H$ which concludes our proof.
