Basis, dense subset and an inequality Let $V \subset H$, where $V$ is separable in the Hilbert space $H$.
So there is a basis $w_i$ in $V$ such that, for each $m$, $w_1, ..., w_m$ are linearly independent and the finite linear combinations are dense in $V$.
Let $y \in H$, and define $y_m = \sum_{i=1}^m a_{im}w_i$ such that $y_m \to y$ in $H$ as $m \to \infty$.
Then, why is it true that $\lVert y_m \rVert_H \leq C\lVert y \rVert_H$?
I think if the $w_i$ were orthonormal this is true, but they're not. So how to prove this statement?

It is not true. Choose $y=0$ and $a_{1,m} = \frac{1}{m}$. Then $y_m \to y$, but it is never the case that $\|y_m\| \leq C \|y\|$.
Elaboration:
This is because $\|y_m\| = \frac{1}{m} \|w_1\|$, the $w_i$ are linearly independent, hence non-zero. Hence $\|y_m\| = \frac{1}{m} \|w_1\| > 0$ for all $m$. There is no choice of $C$ that will satisfy the equation $\|y_m\| \leq C \|y\|$. 
The above is true even if $w_i$ are orthonormal. 
(I think you need to be more explicit about your choice of $a_{im}$. A 'nice' choice would be to let $y_m$ be the closest point to $y$ in $\text{sp}\{w_i\}_{i=1}^m$. This is what the first part of the answer by Berci does below.)
