Showing that the operator $P(v)=\sum_{k\in J}\langle v,e_j\rangle e_j$ is bounded when $(e_k)_{k\in J}$ is an orthonormal set of vectors Let $(e_k)_{k\in J}$ be a collection of orthonormal vectors in a Hilbert space $H$, where $J$ is an index set, and define $P(v) = \sum_{k \in J}\langle v, e_k\rangle e_k$. In the finite case $|J| < \infty$ it is quite clear that $P$ is bounded. However all my intuition regarding the reasoning goes away when we allow the possibility that $J$ has countably infinite or uncountable cardinality. Namely, we know that the ambient Hilbert space $H$ has an algebraic basis $\mathcal{B}$ and therefore given any $v \in H$ we can represent $v$ as a finite linear combinations of elements of $\mathcal{B}$.
But to argue that $P$ is bounded is to argue that either the basis vectors defining $v$ are not orthogonal to only some finitely many members of $(e_k)_{k \in J}$ or that the "components" of the said basis vectors with the elements $e_k, k \in J$, vanish to zero in the (potentially uncountable sum) $\sum_{k\in J}\langle v, e_k\rangle e_k$.
Unfortunately I am not sure how to approach either of these two required branches as I feel that I am having some sort of a chicken and an egg problem. For the record, my functional analysis book has only introduced the orthogonal projection theorem and in principle no mention of orthogonal bases of (any) space has been mentioned yet.
 A: Let $K$ be the Hilbert subspace with orthonormal basis $\{e_k\}_{k \in J}$, that is $K$ is the closed span of the given family of orthogonal vectors. To any closed subspace $K$ of a Hilbert space $H$, there exists a unique orthogonal projections $P_K\in B(H)$ onto this subspace.
We claim that $P_K$ is given by your formula, so it follows that the formula defines a bounded operator.
Indeed, extend $\{e_k\}_{k\in J}$ to an orthonormal basis $\{e_k\}_{k \in I}$ with $J\subseteq I$. Fix $v\in H$. From basic Hilbert space theory, we know that the summation
$$v = \sum_{k \in I} \langle v, e_k\rangle e_k$$
holds where the series converges in the norm-topology. Hence, using continuity of $P_K$ and using that $P_K e_k = 0$ when $k\notin J$
$$P_K v = \sum_{k\in J} \langle v, e_k\rangle e_k.$$

Alternatively, if you don't want to use the correspondence between closed subspaces and orthogonal projections, you can make use of the fact that
$$\sum_{k\in I} \alpha_k e_k$$
converges in norm if and only if $$\sum_{k\in I}|\alpha_k|^2 < \infty.$$
You can apply this to your situation to deduce that $\sum_{k\in J}\langle v, e_k\rangle e_k$ converges (by Bessel's inequality) and then
also $$\left\|\sum_{k\in J} \langle v, e_k\rangle e_k\right\|^2= \sum_{k\in J}|\langle v, e_k\rangle|^2 \le \|v\|^2$$
from which it will also follow that $\|P\| =1$, so $P$ is a bounded operator.
