Let $X$ be an inner product space, let $x \in X$ be non-zero, and let $M$ be an uncountable orthonormal subset of $X$. Then what can we say about the cardinality of the following set? $$ \{ \ v \in M \colon \ \langle x, v \rangle \neq 0 \ \}.$$
I know that, for any orthonormal sequence $(e_k)$ in $X$, the Bessel inequality $$\sum_{k=1}^\infty \vert \langle x, e_k \rangle \vert^2 \leq \Vert x \Vert^2$$ holds and hence, for all $m \in \mathbb{N}$, the the set $$\{ \ k \in \mathbb{N} \ \colon \ \vert \langle x, e_k \rangle \vert > \frac{1}{m} \ \}$$ is at most finite.
But I'm struggling to proceed from the countable case to the uncountable one.
If $M$ is an orthonormal set (the normalisation is not necessary, by the way, that the elements are mutually orthogonal suffices), the set
$$N(x) = \{ v\in M : \langle x,v\rangle \neq 0\}$$
is at most countable.
For a finite subset $F\subset M$, we define
$$x_F := \sum_{v\in F} \langle x,v\rangle\cdot v.$$
Then
\begin{align} 0 &\leqslant \lVert x-x_F\rVert^2\\ &= \langle x-x_F, x-x_F\rangle\\ &= \langle x,x\rangle - \langle x_F,x\rangle - \langle x,x_F\rangle + \langle x_F,x_F\rangle\\ &= \lVert x\rVert^2 - \sum_{v\in F} \bigl\langle \langle x,v\rangle\cdot v,x\bigr\rangle - \sum_{v\in F} \bigl\langle x, \langle x,v\rangle\cdot v\bigr\rangle + \sum_{v\in F} \bigl\langle \langle x,v\rangle\cdot v,x_F\bigr\rangle\\ &= \lVert x\rVert^2 - 2\sum_{v\in F} \lvert\langle x,v\rangle\rvert^2 + \sum_{v\in F} \langle x,v\rangle\langle v,x_F\rangle\\ &= \lVert x\rVert^2 - 2\sum_{v\in F} \lvert\langle x,v\rangle\rvert^2 + \sum_{v\in F} \lvert \langle x,v\rangle\rvert^2\\ &= \lVert x\rVert^2 - \sum_{v\in F} \lvert\langle x,v\rangle\rvert^2. \end{align}
Now, for an arbitrary $c > 0$, consider
$$M_c(x) = \{ v\in M : \lvert\langle x,v\rangle\rvert \geqslant c\}.$$
For any finite subset $F$ of $M_c(x)$, we have by the above
$$c^2\cdot \operatorname{card} F \leqslant \sum_{v\in F} \lvert\langle x,v\rangle\rvert^2 \leqslant \lVert x\rVert^2,$$
and therefore $\operatorname{card} F \leqslant \frac{\lVert x\rVert^2}{c^2}$. It follows that $M_c(x)$ is a finite set with $\operatorname{card} M_c(x) \leqslant \frac{\lVert x\rVert^2}{c^2}$, since otherwise it would have a finite subset with more than $\frac{\lVert x\rVert^2}{c^2}$ elements. Since further
$$N(x) = \bigcup_{m=1}^\infty M_{1/m}(x)$$
exposes $N(x)$ as a union of countably many finite sets, it follows that $N(x)$ is at most countable.
We had shown and used Bessel's inequality
$$\sum_{v\in F} \lvert \langle x,v\rangle\rvert^2 \leqslant \lVert x\rVert^2,\tag{1}$$
for finite subsets $F$ of the orthonormal set $M$ above. Bessel's inequality generalises to arbitrary orthonormal sets, we have
$$\sum_{v\in M} \lvert\langle x,v\rangle\rvert^2 \leqslant \lVert x\rVert^2\tag{2}$$
for every orthonormal set $M$ in an inner product space $H$, and every $x\in H$, but one needs to define the sum of (possibly) uncountably many terms for the left hand side of $(2)$ to make sense.
Generally, that leads to the theory of summable families in abelian topological groups or topological vector spaces, but in the case here, where all terms are non-negative real numbers, a simpler definition suffices, and we can meaningfully assign a sum - either a non-negative real number or $+\infty$ - to every family of non-negative real numbers by defining
$$\sum_{\alpha\in A} a_\alpha := \sup \left\{ \sum_{\alpha\in F} a_\alpha : F\text{ is a finite subset of } A\right\}\tag{3}$$
if $A$ is a set and $a_\alpha$ is a non-negative real number for every $\alpha \in A$.
With this definition, $(2)$ is an immediate consequence of the fact that $(1)$ holds for every finite subset of $M$.
We note that a family $\{ a_\alpha : \alpha \in A\}$ of non-negative real numbers that has a finite sum $S$ can have at most countably many strictly positive members, for the number of members that are not smaller than $\frac{1}{n}$ is bounded by $n\cdot S$ and hence finite. A similar fact is true for summable families in metrisable abelian topological groups or metrisable topological vector spaces, there a summable family can contain at most countably many non-zero members. This does not necessarily hold for summable families in non-metrisable groups or vector spaces, there a summable family can have uncountably many non-zero members.
We further note that definition $(3)$ of the sum of a family of non-negative real numbers is compatible with the familiar definition as the limit of the partial sums for the case of a countable family that is indexed by natural numbers. If $(a_n)_{n\in \mathbb{N}}$ is a sequence of non-negative real numbers, then each partial sum $\sum_{n=0}^k a_n$ of the series $\sum_{n=0}^\infty a_n$ is the sum of a finite subfamily, hence less than or equal to the sum according to definition $(3)$. Conversely, each finite subset $F$ of $\mathbb{N}$ is contained in the initial segment $\{ n : n \leqslant \max F\}$ of $\mathbb{N}$, and therefore
$$\sum_{n\in F} a_n \leqslant \sum_{n=0}^{\max F} a_n \leqslant \lim_{k\to\infty} \sum_{n=0}^k a_n,$$
which shows that the supremum of the sums of finite subfamilies is less than or equal to the limit of the partial sums.
Incidentally, this proves that series with non-negative terms can be reordered arbitrarily without changing the sum, since definition $(3)$ makes no reference to any particular ordering of the family.