Proving that an orthonormal system close to a basis is also a basis Let $\mathcal{H}$ be a Hilbert space and $(e_n)_{n \in \mathbb{N}} \subseteq\mathcal{H}$ be an orthonormal basis and $f_n$ be an orthonormal system such that 
$(f_n)_{n \in \mathbb{N}} \subseteq\mathcal{H}$ and $$\sum_{n \in \mathbb{N}}\|e_n-f_n\|^2 <1.$$ Prove that $f_n$ is also a basis.
I am kind of stuck and can't figure out how to prove that $(f_n)_{n \in \mathbb{N}}$ is a basis.
I tried to prove by contradiction that if $(f_n)_{n \in \mathbb{N}}$ is not  a basis then $\exists   x \neq0 \in \mathcal{H}$ such that $\langle x,f_n \rangle=0 \forall n \in \mathbb{N}$.
I also know that $\|e_n-f_n\|^2 \to 0 \implies Re(\langle e_n,f_n \rangle) \to 1$
I then tried to write $x=\sum_{n=1}^{\infty} \langle x,e_n \rangle e_n$ from which I get that 
$\sum_{k=1}^{\infty} \overline{ \langle x,e_k \rangle} \langle e_k,f_n \rangle \forall n \in \mathbb{N}$
But I do not know how to proceed. Any hints would be highly appreciated
 A: Let $x \in \mathcal{H}$ with $\lVert x\rVert = 1$. Show that $x$ cannot be in the orthogonal complement of $\operatorname{span} \{ f_n : n \in \mathbb{N}\}$.
Since $\{ e_n : n \in \mathbb{N}\}$ is a Hilbert basis of $\mathcal{H}$, we can write
$$x = \sum_{n = 0}^\infty c_n\cdot e_n.$$
Consider now
$$y = \sum_{n = 0}^\infty c_n \cdot f_n.$$
Then we have
$$\lVert x-y\rVert^2 \leqslant \Biggl(\sum_{n = 0}^\infty \lvert c_n\rvert\cdot \lVert e_n - f_n\rVert\Biggr)^2 \leqslant \sum_{n = 0}^\infty \lvert c_n\rvert^2\cdot \sum_{n = 0}^\infty \lVert e_n - f_n\rVert^2 < 1.$$
It follows that
$$\left\lVert x - \sum_{n = 0}^\infty \langle x,f_n\rangle f_n\right\rVert^2 \leqslant \lVert x-y\rVert^2 < 1 = \lVert x\rVert^2,$$
so some $\langle x, f_n\rangle$ must be nonzero.

In fact, we have something much stronger: It already follows that $(f_n)_{n \in \mathbb{N}}$ is also an orthonormal basis when
$$\sum_{n\in \mathbb{N}} \lVert e_n - f_n\rVert^2 < +\infty.\tag{1}$$
Proof: Choose $\varepsilon \in (0,1)$. By $(1)$, there is an $N \in \mathbb{N}$ such that
$$\sum_{n \geqslant N} \lVert e_n - f_n\rVert^2 < \varepsilon^2.$$
Let $E = \overline{\operatorname{span} \{ e_n : n \geqslant N\}}$ and $F = \overline{\operatorname{span} \{ f_n : n \geqslant N\}}$. Define
\begin{gather}
A \colon E^{\perp} \to F^{\perp},\quad x \mapsto x - P_F x,\\
B \colon F^{\perp} \to E^{\perp},\quad x \mapsto x - P_E x,
\end{gather}
where $P_E,\,P_F$ are the orthogonal projections to $E$ resp. $F$.
By $(1)$, we have
$$\lVert P_F x\rVert^2 = \Biggl\lVert \sum_{n \geqslant N} \langle x,f_n\rangle f_n\Biggr\rVert^2 = \sum_{n\geqslant N} \lvert \langle x, f_n\rangle\rvert^2 = \sum_{n\geqslant N} \lvert \langle x, f_n - e_n\rangle\rvert^2 \leqslant \lVert x\rVert^2\sum_{n\geqslant N} \lVert f_n - e_n\rVert^2 < \varepsilon^2\lVert x\rVert^2$$
for all $x \in E^{\perp}$, and similarly
$$\lVert P_E x\rVert^2 = \sum_{n \geqslant N} \lvert\langle x, e_n\rangle\rvert^2 = \sum_{n\geqslant N} \lvert\langle x, e_n - f_n\rangle\rvert^2 < \varepsilon^2\lVert x\rVert^2$$
for all $x\in F^{\perp}$.
For $x \in E^{\perp}$ we have
\begin{align}
BA x &= B(x - P_F x) = (x - P_F x) - P_E(x - P_F x)\\
&= x - P_F x + P_E P_F x = x - (I - P_E)P_F x\\
&= x - P_{E^{\perp}} P_F x
\end{align}
since $P_E x = 0$, and hence $\lVert (I_{E^{\perp}} - BA) x\rVert = \lVert P_{E^{\perp}} P_F x\rVert \leqslant \lVert P_F x\rVert < \varepsilon \lVert x\rVert$. It follows that $BA$ is an automorphism of $E^{\perp}$, in particular that $A$ is injective, and hence $\dim E^{\perp} \leqslant \dim F^{\perp}$.
In the same way, for $x \in F^{\perp}$ we have
\begin{align}
AB x &= A(x - P_E x) = (x - P_E x) - P_F(x - P_E x)\\
&= x - P_E x + P_F P_E x = x - (I - P_F)P_E x\\
&= x - P_{F^{\perp}} P_E x,
\end{align}
and consequently $\lVert I_{F^{\perp}} - AB\rVert < \varepsilon$. So $AB$ is an automorphism of $F^{\perp}$, in particular $B$ is injective and hence $\dim F^{\perp} \leqslant \dim E^{\perp}$.
Since $(e_n)_{n \in \mathbb{N}}$ is by assumption a Hilbert basis of $\mathcal{H}$, it follows that $(e_n)_{n < N}$ is a basis of $E^{\perp}$, whence $\dim E^{\perp} = N$ (we assume $0 \in \mathbb{N}$, the dimension would be $N-1$ otherwise). By the orthonormality of $(f_n)_{n\in\mathbb{N}}$, the $N$-element family $(f_n)_{n < N}$ is linearly independent and contained in $F^{\perp}$. Since $\dim F^{\perp} = N$, it is a basis of $F^{\perp}$, and thus $(f_n)_{n \in \mathbb{N}}$ is total in $\mathcal{H}$.
