This subspace of $L^2$ is finite dimensional?

Question

For the Lebesgue measure let F $ \subset L^2( [0,1], \mathbb{R} ) $ be a closed linear subspace of $ L^2 $ and also $ F \subset (\mathcal{C} [0,1]) $.

With the Banach isomorphism theorem i proved that $ (F, \| \|_{ L^2} ) \cong (F, \| \|_{ \infty} ) $ and so both norms are equivalent on F.

3) Suppose that $ ( \varphi_n)_{1 \leq n \leq N} $ is a orthonormal subset in $ (F, \| \|_{ L^2} ). $ For all $ y \in [0,1] $ we consider ,

$ f_y : x \in [0,1] \mapsto \sum_{n=1}^N \varphi_n(x) \varphi_n(y) $

a) for $ y \in [0,1] $ compute $ \| f_y \|_{ L^2} . $

b) for $ y \in [0,1] $ find a lower bound to $ \| f_y \|_{ \infty} . $

4) Conclude that F is finite dimensional.

So i found $ \| f_y \|_{ L^2} = (\sum_{n=1}^N \varphi_n^2(y))^{1/2} $

And $ \varphi_i (y) \leq \| f_y \|_{ L^2} \leq \| f_y \|_{ \infty} . \forall i $

---

How do i conclude that F is finite dimensional ?

I tried using the Riesz Lemma (compactness of unit ball thus finite dimensional) without any success. I know that $ L^2 $ is a separable Hilbert Space hence it has a countable Hilbert basis but i'm not sure if that helps here.

Thanks in advance for any help on the matter.