In functional analysis, we have introduced Schauder and Hamel Bases, where a Hamel Bases is "stronger" in the sense that it can generate an entire space with only finitely many linear combinations per element that is. While a Schauder Basis requires countably many linear combinations.
Now, we have defined for a Hilbert space $\mathcal{H}$, an orthonormal basis $\{e_{\alpha}\}_{\alpha \in J}$ where $J$ is an index set, so that:
$i) \langle e_{\alpha}, e_{\beta}\rangle=\delta_{\alpha \beta}$
$ii) \langle e_{\alpha}, f\rangle=0$ for all $\alpha \in J\Rightarrow f=0$
We also have some useful properties, like Parseval's Equality (Bessel's Inequality), i.e. $\vert \vert x \vert \vert ^{2}=\sum\limits_{\alpha\in J}\vert \langle e_{\alpha},x\rangle\vert^{²}$.
But nowhere in the above is actually stated that $\{e_{\alpha}\}_{\alpha \in J}$ can be viewed as an "actual" basis, that is in my understanding, we can represent any $x \in \mathcal{H}$ as: $x = \sum\limits_{\alpha \in J}c_{\alpha} e_{\alpha}$.
I know that in finite dimensions, we can represent any $x \in \mathcal{H}_{\operatorname{finite-dimensional}}$ as $x=\sum\limits_{j=1}^{N}\langle e_{j},x\rangle e_{j}$, so I am assuming the above does not hold for infinite dimensions.
In our lectures, we have also stated that a countable orthonormal basis $\{ e_{n}\}_{n \in \mathbb N}$ of a Hilbert Space $\mathcal{H}$ can also be viewed as a Schauder Basis, so that poses the following question:
Since for any $x \in \mathcal{H}$ I can find a sequence $(x_{n})_{n} \subseteq \mathbb R$ where $\vert \vert x - \sum \limits_{n =1}^{N}x_{n}e_{n} \vert \vert\xrightarrow{N \to \infty}0$, can I then simply represent $x = \sum\limits_{n \in \mathbb N}x_{n}e_{n}$?
If an orthonormal set $\{e_\alpha:\alpha\in\mathcal{A}\}$ is maximal (that is, it is not contained in a larger orthonormal set), then we say that it's an orthonormal basis. This is actually equivalent to its linear span being dense in $H$. Further, for every $f\in H$, we have that $$f=\sum\limits_{\alpha\in \mathcal{A}}(f,e_\alpha)e_\alpha.$$ Here, this convergence is understood in the norm topology on $H$. You can compare the consistency of this decomposition directly to the second condition of your definition of orthonormal basis. I can sketch a proof of the decomposition, if needed.
EDIT: I'm just going to talk about separable spaces (since for a fixed $f$, we can only have that $(f,e_\alpha)\neq 0$ for (at most) countably many $\alpha\in\mathcal{A}$). So, let me consider $\{e_j: j=1,2,\cdots\},$ relabeling if necessary. Consider the orthogonormal projection onto $L_k=\text{span}\{e_j: 1\leq j\leq k\}$ $$P_k f=\sum\limits_{j=1}^k(f,e_j) e_j.$$ Note that for any $k$, $$\|f\|^2=\|P_k f\|^2+\|f-P_k f\|^2\geq\|P_k f\|^2=\sum\limits_{j=1}^k |(f,e_j)|^2.$$ Since this is true for any $k$, the infinite series converges. We also can see that if $n\geq k,$ then $$\|P_n f-P_k f\|^2=\sum\limits_{j=k+1}^n |(f,e_j)|^2,$$ implying that $(P_k f)$ is Cauchy, for each $f$. In order to see that the limit is $f$, note that $$g=\sum\limits_{j=1}^\infty (f,e_j)e_j$$ is an element of $H$. Observe further that $(f-g,e_j)=0$ for each $e_j$, and hence with each element of the span, and hence with all elements of the closure; it follows that $f-g=0$.