Construct a strictly increasing function $\varphi: \mathbb{N}\to \mathbb{N}$ Let $(x_n)_{n}\subseteq E$ and $(y_n)_{n}\subseteq E$ such that $\|x_n\|=\|y_n\|=1$, where  $E$ is a complex inner product space.
Consider the following property $(P)$:
$(P)$: There exists a constant $c< 1$ such that $$|\langle x_n, y_n\rangle|\leq c< 1,\;\forall n\in \mathbb{N}.$$

If $(P)$ fails to hold, why there exists a strictly increasing function $\varphi: \mathbb{N}\to \mathbb{N}$ such that
  $$\displaystyle\lim_{n\longrightarrow\infty}|\langle x_{\varphi(n)}\; ,\;y_{\varphi(n)}\rangle|=1.$$

My attempt: If $(P)$ fails to hold, then $\forall\, c<1$, there exists $n_c\in \mathbb{N}$ such that $|\langle x_{n_c}\; ,\;y_{n_c}\rangle|> c$.
If $c=\frac{n}{n+1}<1$, then $|\langle x_{n_c}\; ,\;y_{n_c}\rangle|> \frac{n}{n+1}$. On the other hand by Cauchy-Shwartz inequality
$$|\langle x_{n_c}\; ,\;y_{n_c}\rangle| \leq \|x_{n_c}\|\|y_{n_c}\|=1.$$
My problem is how to construct explicitly the function $\varphi$
Note that this problem is taken from this note (Theorem 7.4. page 21)

 A: It is not true: if there is $N$ such that $|\langle x_N, y_N\rangle| =1$ and $|\langle x_n, y_n\rangle| =0$ for all $n\neq N$, then $(P)$ fails but for any strictly increasing $\varphi$ we have $\lim_{n\to\infty}|\langle x_{\varphi(n)}, y_{\varphi(n)}\rangle| =0$.
However, we can construct $\varphi$ inductively if for all $c\in (0,1)$, there exist infinitely many $n$ such that
$$
|\langle x_n, y_n\rangle| > c.
$$ Assume we have constructed strictly increasing $\varphi(j)$ for $j\le n$ for which $|\langle x_{\varphi(j)}, y_{\varphi(j)}\rangle| > 1-\frac{1}{2^j}$ holds. Since there are infinitely many $k$ such that
$$
|\langle x_k, y_k\rangle| >1-\frac{1}{2^{n+1}},\tag{*}
$$ we can find $k>\varphi(n)$ such that $(*)$ holds. By letting $\varphi(n+1)=k$, we have $\varphi(j)$ for $j\le n+1$ such that $|\langle x_{\varphi(j)}, y_{\varphi(j)}\rangle| > 1-\frac{1}{2^j}$ holds. By induction, $\varphi$ can be defined on $\mathbb{N}$ and it follows
$$
\lim_{j\to\infty}|\langle x_{\varphi(j)}, y_{\varphi(j)}\rangle|=1
$$ from the construction.
EDIT: In view of the lecture note, the correct version of $(P)$ is: for some $c\in (0,1)$, $$
|\langle x_n, y_n\rangle|\leq c< 1$$ for all sufficiently large $n$. If we negate this statement, it follows that for all $c\in (0,1)$, $$\langle x_n, y_n\rangle|>c$$ for infinitely many $n$. Hence, the above construction works and gives the existence of a subsequence $\varphi(n)$ such that $$\lim\limits_{n\to\infty}|\langle x_{\varphi(n)}, y_{\varphi(n)}\rangle|=1.$$
