$\textbf{Theorem:}$ Let $X$ a inner product space and $\{e_n\}_{n \in\mathbb{N}}$ a orthonormal basis of $X$. (Is orthonormal and $\overline{span\{e_n\}} = X $) Prove that for each $x\in X$
$$ x = \sum_{i=1}^{\infty} \langle x,e_i \rangle e_i $$
$\textbf{My attempt:}$ I use this lemma:
$\textbf{Lemma:}$ Let $\{e_1,e_2, \ldots,e_n \}$ a orthonormal set. Then, for each $\lambda_1, \ldots,\lambda_n \in \mathbb{C}$
$$ \Vert x - \sum_{i=1}^{n} \langle x,e_i \rangle e_i \Vert \leq \Vert x - \sum_{i=1}^{n} \lambda_i e_i \Vert $$
Now, $x\in X$ and $\epsilon >0$. By density exists $\alpha_1, \ldots, \alpha_k$ such that
$$ \Vert x - \sum_{i=1}^{k} \alpha_i e_i \Vert < \epsilon $$
By lemma we have for $m \geq k$ ($\alpha_i=0, i>k$)
$$ \Vert x - \sum_{i=1}^{m} \langle x,e_i \rangle e_i \Vert \leq \Vert x - \sum_{i=1}^{k} \alpha_i e_i - \sum_{i=k+1}^{m} 0 e_i \Vert = \Vert x - \sum_{i=1}^{m} \alpha_i e_i \Vert = \Vert x - \sum_{i=1}^{k} \alpha_i e_i \Vert < \epsilon $$
So
$$\Vert x - \sum_{i=1}^{m} \langle x,e_i \rangle e_i \Vert \to 0 $$
Hence
$$x = \sum_{i=1}^{\infty} \langle x,e_i \rangle e_i $$
My proof is correct?
Well, yes: your proof looks fine...but I think it is way easier. For any $\;x\in V\;$ write
$$x=\sum_{n=1}^\infty a_ie_i\implies \forall\,k\in\Bbb N\,,\;\color{red}{\langle x,\,e_k\rangle}=\langle \sum_{n=1}^\infty a_ie_i,\,e_k\rangle=\sum_{n=1}^\infty a_i\langle e_i,\,e_k\rangle=\color{red}{a_k}$$
and there you are...