Is there a more elementary proof of this Hilbert Space fact?

Question

Suppose $H$ be a Hilbert space in which there is an $A=\left \{ y_i \right \}^{\infty }_{i=1}$ such that any vector $x\in H$ is a $finite$ linear combination of elements from $A$. Then $H$ is finite dimensional. The result is true for an arbitrary Banach space as can be seen by defining $A_k=$ span $\left \{ y_1,\cdots ,y_k \right \}$ and noting that by hypothesis $H=\bigcup _kA_k$ and then the result follows by a typical Baire argument. The problem is, the exercise appears in a book before Baire's Theorem is proved, so I am looking for a more elementary proof, one that uses specifically the fact that $H$ is a Hilbert space.

Answer 1

If you apply Gram-Schmidt to $A$, you'll get an orthonormal set $\{z_n\}_{n=1}^\infty$ such that every element of $H$ is a finite linear combination of it. If the set $\{z_n\} $ is infinite we get a contradiction, because the orthonormality implies uniqueness of coefficients, and it is impossible to represent, say $\sum_n\frac1n\,z_n$, as a finite sum.