Show there exists a Cauchy subsequence Let $X$ be a separable reflexive real Banach space and $\{\psi_n\}$ be a dense sequence in 
$$\{\psi\in X' : ||\psi||_{X'} \leq 1\}.$$
Consider in $X$ the scalar product defined by 
$$(x | y)_0 = \sum_{n=1}^\infty 2^{-n} \langle \psi_n,x \rangle\langle \psi_n,y \rangle.$$
Show every bounded sequence in $X$ admits a Cauchy subsequence with respect to the norm $||\cdot||_0 $ (norm induced by $(x| x)_0$) .
Proof: 
Since $X$ is reflexive, by Banach–Alaoglu theorem, every bounded sequence $\{x_j\}$ in $X$, there exists a weakly convergent subsequence $\{{x_j}_k\}$ to some $x$. 
To show that ${{x_j}_k}$ is Cauchy under $||\cdot||_0$, it is sufficient to show that the sequence ${{x_j}_k}$ converges to $x$ under  $||\cdot||_0$.
Observe that if ${x_j}_k \rightharpoonup x$, then $||{x_j}_k||_X \leq C$ and 
$$|\langle \psi_n,{{x_j}_k} - x \rangle| \leq ||\psi_n||_{X'}||{{x_j}_k} - x ||_X \leq 2C$$
Now let $\epsilon$ be given, 
$$||{x_j}_k - x||_0^2 = \sum_{n=1}^\infty 2^{-n} {\langle \psi_n,{x_j}_k - x \rangle }^2$$
we split the sum into two parts at $N$ such that 
$$\sum_{n=N}^\infty 2^{-n} {\langle \psi_n,{x_j}_k - x \rangle }^2\leq \sum_{n=N}^\infty 2^{-n} (2C)^2 \leq \epsilon/2.$$
Now for the first $N-1$ terms, choose $K$ such that for $k\geq K$ we have
$$\sum_{n=1}^{N-1} 2^{-n} {\langle \psi_n,{x_j}_k - x \rangle }^2 \leq \epsilon/2,$$
the reason we could choose such $K$ is because $\langle \psi_n,{x_j}_k - x \rangle$ goes to zero as $k$ goes to $\infty$ for each of the $N-1$ terms.
Combine the two, we have for each $k\geq K$
$$||{x_j}_k - x||_0^2 = \sum_{n=1}^{N-1} 2^{-n} {\langle \psi_n,{x_j}_k - x \rangle }^2+ \sum_{n=N}^\infty 2^{-n} {\langle \psi_n,{x_j}_k - x \rangle }^2\leq  \epsilon/2 + \epsilon/2.$$
Questions:


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*Is my proof correct? Is there an easier way to do this? I know my proof is quite long.. Thank you for reading it!

*What is the significance of $\psi_n$ being dense? I did not use this fact in my proof.

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*It looks fine.

*I think the denseness is used implicitly when you want to show that $\|\cdot\|_0$ is a norm.

