In what follows we prove that:
If $\{\lambda_n\}$ are eigenvalues (corresponding to linearly independent eigenvectors) of the self-adjoint compact operator $K$, then $\lambda_n\to 0$.
Suppose not, and in particular, that there exists an $\varepsilon>0$, such that
$\lvert\lambda_{j_n}\rvert\ge\varepsilon$, for $\{\lambda_{j_n}\}$ a subsequence of $\{\lambda_{n}\}$, and let $u_n$ be unit corresponding eigenvectors, i.e., $Ku_n=\lambda_{j_n}u_n$. As $K$ is compact and $\{u_n\}$ is a bounded sequence, then $Ku_n$ has a convergent subsequence. Rename this subsequence as $Kv_n=\mu_nv_n$.
As $K$ is self-adjoint, we may assume that $\langle v_i,v_j\rangle=\delta_{ij}$.
Assume that $Kv_n\to w$, i.e., $\mu_nu_n\to w$. Clearly $$\lvert\mu_n\rvert=\lvert\mu_nv_n\rvert\to \|w\|,
$$
and as $\lvert\mu_n\rvert\ge\varepsilon$, then $\|w\|\ge \varepsilon $.
Meanwhile,
$$
\langle Kv_j,v_n\rangle=\langle \mu_jv_j,v_n\rangle=\mu_j\langle v_j,v_n\rangle=0,
$$
for $n>j$, and thus
$$
\langle w,Kv_n\rangle=\mu_n\langle w,v_n\rangle=0,
$$
for all $n\in\mathbb N$. Thus
$$
\|Kv_n-w\|^2=\langle Kv_n-w,Kv_n-w\rangle=\|Kv_n\|^2+\|w\|^2-2\mathrm{Re}\,\langle Kv_n,w\rangle=\|Kv_n\|^2+\|w\|^2\ge \varepsilon^2,
$$
which is a contradiction.
I don't know what "unique under change of orthonormal basis" means but here is what we can say about uniqueness. If $T$ is a compact self-adjoint operator, then there exists a unique set of pairs $\{(\lambda_i,P_i)\}_{i}$ such that each $\lambda_i$ is a nonzero scalar, the $\lambda_i$ are all distinct, each $P_i$ is a nonzero projection operator, the $P_i$ are pairwise orthogonal, and $$T=\sum_i \lambda_iP_i.$$ Moreover, the $\lambda_i$ are exactly the nonzero eigenvalues of $T$, $P_i$ is the projection onto the eigenspace of $T$ with eigenvalue $\lambda_i$, and each $P_i$ has finite rank.
For the second representation $$Th=\sum_n \mu_n (h,e_n)e_n,$$ the only uniqueness statement we can say is that coefficients $\mu_n$ are uniquely determined (up to permutation) by $T$. Indeed, these coefficients $\mu_n$ are none other than the eigenvalues of $T$, listed with multiplicity. This formula should not look mysterious at all: if we were in a finite-dimensional space and $\{e_n\}$ was the standard (finite) basis, this would literally just be saying that $T$ is the diagonal matrix with the $\mu_n$ as the diagonal entries. So this formula is saying that there exists an orthonormal basis in which $T$ is a diagonal matrix. As usual for diagonal matrices, the diagonal entries are just the eigenvalues (with multiplicity). The orthonormal basis $\{e_n\}$ with respect to which $T$ is diagonal is not unique, just as a diagonalizable matrix does not have a unique basis of eigenvectors.
To relate the two representations, note that if $e_{i_1},\dots,e_{i_k}$ is an orthonormal basis for the eigenspace of $T$ with eigenvalue $\lambda_i$, then the projection $P_i$ onto that eigenspace is just $$P_ih=\sum_{j=1}^k (h,e_{i_j})e_{i_j}.$$ The second representation is obtained from the first representation by just picking an orthonormal basis for each eigenspace in this way and then using this formula for $P_i$. The non-uniqueness of the second representation comes from the fact that there are many different orthonormal bases for each eigenspace.
Best Answer
Let $\mu$ denote the projection valued measure associated to $A$. Note that for any two sets $X,Y\subset\Bbb R$ you have that $\mu(X)$ and $\mu(Y)$ are commuting projections. You can assume by the spectral theorem $A$ is of the form $f\mapsto (x\mapsto x\cdot f(x))$ on $L^2([0,1],d\lambda)$ for some measure $\lambda$, this will help you visualise whats happening.
Note that $A-\frac12\mu([\frac12,1])$ then corresponds to multplication with $x-\frac12 \chi_{[\frac12,1]}(x)$, ie it is $x$ until value $1/2$ and then reset to $0$ after which it grows to $1/2$ again.
Do this visualisation again, what function grows to $\frac14$ and then resets to $0$, then grows to $\frac14$ and resets and so on? The function would be $$x-\left(\frac12\chi_{[\frac12,1]}(x) + \frac14\chi_{[\frac14,\frac12)}(x) + \frac14\chi_{[\frac34,1]}(x)\right)$$ now for the next step you want it to reset in intervals of $\frac18$, so you will need to add some more projections. Continue in this way, being more explicit you get that $$P_1=\mu([\frac12,1]), \quad P_2=\mu([\frac14,\frac12)\,\cup\,[\frac34,1]) \\ P_3=\mu([\frac18,\frac14)\,\cup\,[\frac38,\frac12)\,\cup\,[\frac58,\frac34)\,\cup\,[\frac78,1])$$ and so on, it can be viewed as an exercise to get the explicit forms of the sets you are interested in.