We are in $L^2([0;1];\mathbb{R})$. Let $F$ be a closed subspace of $L^2([0;1];\mathbb{R})$ which verify that for each function $f \in F$, $f \in C^0([0;1])$. We have already proved that $||.||_2$ and $||.||_\infty$ are equivalent norm of $F$. Moreover, if we note $(e_i)_{i \in I}$ a Hilbert basis of $F$, we have proved that :
$\forall x \in [0;1] \quad \exists K_x > 0 \quad \sum_{i \in I} |e_i(x)|^2 \leq K_x$
And I have to deduce that $F$ is a finite dimensional space.
I wanted to prove that $I$ is finite, and then the proof will be ended. To do that, I imagine that I have to use the hypothesis that $e_i$ is continuous and $||e_i||_2 = 1$. But I tried several things and nothing have worked yet.
Someone could help me ? :)
Because $||\cdot||_2$ and $||\cdot||_\infty$ are equivalent, there exists $C>0$ such that $||e_i||_\infty\leq C||e_i||_2=C$ holds for all $i\in I$. By definition, this means $|e_i(x)|\leq C$ holds for all $i\in I$ and $x\in[0,1]$.
For $i\in I$ define $U_i:\{x\in[0,1]:|e_i(x)|\geq\frac12\}$. Then $$||e_i||_2= \sqrt{\int_0^1|e_i(x)|^2dx}= \sqrt{\int_{U_i}|e_i(x)|^2dx+\int_{U_i^c}|e_i(x)|^2dx}\leq \sqrt{C^2\lambda(U_i)+\frac14\lambda(U_i^c)}.$$ Here $\lambda$ represents the Lebesgue measure. Notice that $\lambda(U_i^c)=1-\lambda(U_i)$. Since $||e_i||_2=1$, it follows that $$C^2\lambda(U_i)+\frac14(1-\lambda(U_i))\geq1,\\ (C^2-\frac14)\lambda(U_i)\geq\frac34.$$ Notice that this means that $C^2>\frac14$. Therefore, we find a lower bound $L$ on $\lambda(U_i)$.
Assume for the contrary that $I$ is infinite. Your result: $$\forall x\in[0,1]:\sum_{i\in I}|e_i(x)|^2<\infty$$ implies that $\limsup U_i=\emptyset$. However, we also have $\limsup \lambda(U_i)\geq L>0$. This is a contradiction (Measure and limsup and liminf) so $I$ must be finite.