Hilbert-Schmidt Theorem says that suppose $A$ is a self-adjoint compact operator on the Hilbert space $X$, then $X$ has an orthonormal basis $\{e_i \,|\, i \in I\}$ ($I$ is the index set) which is composed of the eigenvectors of $A$. Therefore, for any $x \in X$, we have \begin{gather} x=\sum_{i \in I} (x,e_i) e_i, \quad \quad (1) \\ Ax=\sum_{i \in I} \lambda_i (x,e_i) e_i, \quad \quad (2) \end{gather} where $\lambda_i$ is the eigenvalue corresponding to $e_i$ ($\lambda_i$ may equal $\lambda_j$ even if $i \neq j$, and the set of eigenvalues of $A$ is countable even if $\{e_i\}$ is not). Through Bessel's inequality, there are countable $(x,e_i)$ not equal to $0$, thus the summations above are actually over countable sets. Then it is said in the textbook that if we list the eigenvalues of $A$ as follows: \begin{equation} |\lambda_1| \geq |\lambda_2| \geq \cdots, \quad \quad (3) \end{equation} then according to Hilbert-Schmidt Theorem, for any $x \in X$, we have \begin{equation} \left\|\left(A-\sum_{i=1}^n \lambda_i e_i \otimes e_i\right)(x)\right\| = \left\|Ax-\sum_{i=1}^n \lambda_i (x,e_i) e_i\right\| = \left\|\sum_{i=1}^\infty \lambda_i (x,e_i) e_i\right\| = \left(\sum_{i=n+1}^\infty \lambda_i^2 |(x,e_i)|^2\right)^{1/2} \leq |\lambda_{n+1}| \left(\sum_{i=n+1}^\infty |(x,e_i)|^2\right)^{1/2} \leq |\lambda_{n+1}| \|x\|, \quad \quad (4) \end{equation} where $e_i \otimes e_i: X \to X; \ x \mapsto (x,e_i) e_i$ is the rank $1$ operator, therefore we have \begin{equation} \left\|A-\sum_{i=1}^n \lambda_i e_i \otimes e_i\right\| \leq |\lambda_{n+1}| \to 0 \quad (n \to \infty), \quad \quad (5) \end{equation} that is, \begin{equation} A=\sum_{i=1}^\infty \lambda_i e_i \otimes e_i, \quad \quad (6) \end{equation} which implies that a self-adjoint compact operator on a Hilbert space can be approximated by the linear combination of rank $1$ operators.
So here comes the question, when $\{e_i\}$ is uncountable, although there are countable $(x,e_i)$ not equal to $0$, it seems that which of them are not equal to $0$ depends on $x$, and accordingly, the addends in the summation $\sum_{i=1}^n \lambda_i e_i \otimes e_i$ in (4) actually vary with $x$, so why can we still derive (5) from (4)?
The addends in the summation $\sum_i \lambda_i (e_i \otimes e_i)$ do not depend on $x$. Indeed, $\{e_i\}_{i \in I}$ and $\{\lambda_i\}_{i \in I}$ were introduced before we even spoke about a fixed element $x \in X$. Hence, as an abstract element, the operator $\sum_i \lambda_i (e_i \otimes e_i)$ does not depend on $x$ in any way.
Therefore, $(5)$ follows from $(4)$ by definition of the operator norm. More generally, the proof uses that if $T: X \to X$ is a linear map with $\|Tx\| \le M\|x\|$ for all $x \in X$, then $\|T\| \le M$. Here, $$T= \sum_i \lambda_i (e_i \otimes e_i).$$ Hopefully this clears your confusion.