Brezis 5.28.2 About the competition of an orthogonal sequence

Question

Let $(e_n)$ be an orthonormal sequence in $H$ is parable. Prove that there exists an orthonormal basis of $H$ that contains $\bigcup_{n\geq 1} \{e_n\}$

My try:

Define the set $\Omega$ as follows:

$$\Omega = \{X \subseteq H : \{e_1\} \subseteq X \text{ and } X \text{ is an orthonormal sequence of } H\}$$

Now, we can proceed with the proof:

1. Define the Partial Order: Let the partial order on $\Omega$ be set inclusion, i.e., $X \leq Y$ if $X \subseteq Y$ for $X, Y \in \Omega$.

2. Show Non-Emptyness: Prove that $\Omega$ is non-empty. This follows from Hipotesis.

3. Show Chains Have Upper Bounds: Take an arbitrary chain $\mathcal{C}$ in $\Omega$, and consider the union of all sets in this chain, i.e., $\bigcup \mathcal{C}$ is an orthonormal sequence size is partial ordened and $e_1$ is on all elements.

4. Apply Zorn's Lemma: By Zorn's lemma, there exists a maximal element $M$ in $\Omega$.

5. Show $M$ is a Basis of $H$: Do by contradicción.

6. Prove $M$ Contains the Given Sequence: Since $\{e_1\} \subseteq M$ (as per the definition of $\Omega$), and $M$ is maximal, it follows that $M$ contains the sequence $\{e_n\}$.

is fine ?