Alternative Proof for a Lemma in Barry Simon's "Operator Theory" I've been working on a presentation based on chapters $3.6$, $3.7$ of Barry Simon's book "Operator Theory". Due to having only a limited amount of time to present the topic, I cannot introduce all the notions from the previous chapters of his book. This results in problems with some of the proofs in the book, which I had to change so as not to include some of these notions.  There is one lemma at the beginning of chapter $3.6$, which I can't prove though: 

$\mathbf{3.6.3.Lemma:}$ Let $A\in L(H)$ (where L(H) is the set of BLOs over a Hilbert space H) satisfy for all Orthonormal Systems $(\phi_n)_{n\in\mathbb{N}},(\psi_n)_{n\in\mathbb{N}}$: $${\langle\psi_n,A\phi_n\rangle}\overset{n\rightarrow +\infty}{\longrightarrow}0$$
  Then $A$ is compact.

Barry Simon uses the concept of the Polar Decomposition to prove this with the help of another Lemma from a previous chapter ($3.1.14$) and his proof is pretty straightforward. However, the people attending the presentation don't know what a Polar Decomposition is, nor do they have the prerequisite knowledge for me to quickly introduce the concept (This would require them to know about projections and partial isometries). Additionally, Lemma $3.1.14$ which is used in this proof seems to be a special case of what we are trying to prove. 
Thus I'm trying to prove this using only basic properties of compact operators, such as Schauder's Theorem, the fact that the limit of a sequence of finite rank compact operators is compact and the equivalent definitions for compactness. I'm also allowed to use the canonical decomposition and singular values. What I can also use is the property that the Eigenvalues of compact operators converge to zero. The basic difficulty I'm facing is that this lemma is unrelated to the topic I'm presenting and it requires knowledge, with which I'm not very familiar.
I have attempted to prove it by approximating it by a sequence of finite rank operators, the problem is that I can't think of a sequence, which takes advantage of the assumption. Using the canonical decomposition does not work since it requires that the operator be compact (so I can only use it in a proof by contradiction). So I don't have any idea how to start.
Another approach I have tried is proving the contraposition using the property of the Eigenvalues. Basically, assuming that the Eigenvalues (so also the singular values) converge to something other than $0$ (which would imply that $A$ is not compact) then the inner product with $\phi$ and $\psi$ does not always converge to zero, but I haven't got far. 
So I would appreciate it if any of you could give me some assistance in proving this (hints, ideas etc.) 
-Thanks in advance!  
 A: $\newcommand{\N}{\mathbb{N}}$
Let $(x_n)_{n\in\N}$ be a bounded sequence, I want to show that $(A x_n)_{n\in\N}$ has a convergent subsequence. By Banach-Alaoglu, $(x_n)_{n\in\N}$ has a weakly convergent subsequence, so wlog let $(x_n)_{n\in\N}$ be a weak null sequence, i.e. $x_n\rightharpoonup 0$. Up to a subsequence, using Gram-Schmidt we can find an orthogonal sequence $(\phi_n)_{n\in\N}$ such that
$$
\|x_n-\phi_n\|_H<\frac{1}{n}.
$$For details, see Lemma 3 in the first answer to this question (thanks to MrVolt16).
By assumption, you have that $A\phi_n\rightarrow 0$ strongly*, hence
$$
\|Ax_n \|_H \leq \|A\phi_n\|_H+\|A(x_n-\phi_n)\|_H\rightarrow0
$$and consequently $A x_n\rightarrow 0$ up to subsequences. This proves that $A$ is sequentially compact and hence compact.
EDIT:
*As pointed out in the comments, $A\phi_n\rightarrow 0$ strongly is indeed not as obvious as I thought, here is a possible fix, up to subsequences:
Since $A\phi_n$ is still bounded, we again may assume $A\phi_n\rightharpoonup 0$ up to subsequences and similar to before introduce an orthogonal sequence $(\psi_n)_{n\in\N}$ such that
$$
\|A\phi_n-\psi_n\|_H<\frac{1}{n}
$$up to subsequences.
Using this sequence, we may calculate
$$
\|A\phi_n\|_H^2=\langle A\phi_{n},A\phi_{n}\rangle_{H}
=\langle\psi_{n}+A\phi_{n}-\psi_{n},A\phi_{n}\rangle_{H}$$
$$
=\langle\psi_{n},A\phi_{n}\rangle_{H}+\langle A\phi_{n}-\psi_{n},A\phi_{n}\rangle_{H}\rightarrow 0
$$ for $n\rightarrow\infty$, the first term by assumption and the second term by construction of $\psi_n$, boundedness of $A\phi_n$ and Cauchy-Schwartz. Hence $A\phi_n\rightarrow 0$ strongly up to subsequences.
