A similar proposition to Riesz's lemma 
Let $H$ be a separable Hilbert space, $\{u_k\}$ is a non precompact sequence with $\|u_k\|=1$ , then there exists a subsequence $\{v_k=u_{n_k}\}$ and a constant $a>0$ such that
  $$d(v_k,\mathrm{span}\{v_1,\cdots,v_{k-1}\})>a$$

I think it's similar to Riesz's lemma, but I do not have idea to use the condition 'non precompact', any help will be appreciated.
 A: First of all, you don't need that space $H$ is separable (it is obvious since your statement is about sequence $\{u_n\}$ that belongs to a subspace $H_0 = \overline{Span \{u_n\}}$ and $H_0$ is a separable Hilbert space). So I will not use this assumption (that was a note that explains why this assumption is not necessary).
Next we observe that if a metric space $X$ is not precompact (I suppose that precompact is a synonym for totally bounded) then you can construct a number $\alpha_0 > 0$ and a sequence $x_n \in X$ such that $d(x_n, x_m) > \alpha_0$ for all $n \ne m$. So we choose a subsequence $w_k = u_{n_k}$ such that $||w_n - w_m|| > \alpha_0$ for some $\alpha_0 > 0$ and $n \ne m$. 
Now we construct a sequence $v_k = w_{m_k}$ by induction for arbitrary $a < \frac{\alpha_0}{2}$. $v_1$ is set to be $w_1$. Next, assume that we have already constructed $v_1, \dots, v_k$. If you can't make a choice then for all $m > m_k$ you have $d(w_m, Span\{v_1, \dots, v_k\}) \le a = \frac{\alpha_0}{2}$. Let $w_m = h_m + r_m$ where $h_m$ is orthogonal to $Span\{v_1, \dots, v_k\}$ and $r_m \in Span\{v_1, \dots, v_k\}$. So, $||h_m|| \le a$ and  $1 \ge ||r_m||^2 = 1 - ||h_m||^2 \ge 1 - a^2$. Then $||w_m - w_n||^2 = ||r_m - r_n||^2 + ||h_m - h_n||^2 \le ||r_m - r_n||^2 + 4a^2 < ||r_m - r_n||^2  + \alpha^2_0$. Now we observe that $r_m$ is a bounded sequence in finite dimensional space and therefore value $||r_m - r_n||$ can be made arbitrarily small for $m \ne n$; $m,n > m_k$ and this contradicts with assumption that $||w_n - w_m|| > \alpha_0$ for all $m \ne n$.
