Fredholm alternative and orthonormal basis The following question relates to the Fredholm alternative:
Let $K:H \rightarrow H$ be a compact linear operator and let $I$ be the 
identity operator. 
Notation: $N$ is the nullspace and $R$ is the range.
1.The proof of the first part '$N(I-K)$ is finite dimensional' states the following: If we suppose that $\text{dim}N(I-K) = +\infty$. Then we can find a countably infinite orthonormal basis $\{u_{k}\}_{k=1}^{\infty} \subset N(I-K)$ where $(u_{i},u_{j}) = \delta_{i,j}$.
How are we guaranteed to find such a orthonormal basis for an infinite dimensional subspace? What result is used? Gram-Schimdt only applies to finite-dimensional spaces if I'm not mistaken.
2.The following is taken from a the proof of another property of Fredholm Alternative proof:
Let us consider $\{v_{k}\}_{k}^{\infty} \subset R(I-K)$ where $v_{k} \rightarrow v$. Then, there exists $x_{k} \in H$ which solve $(I-K)x_{k} = v_{k}$. Since $x_{k}$ can be expressed as a sum of basis vectors from $N(I-K)$ and $N(I-K)^{\perp}$, then we can say there exists $u_{k} \in N(I-K)^{\perp}$ solving $u_{k} - Ku_{k} = v_{k}$.
How does it follow that $x_{k}$ can be expressed as a sum from $N(I-K)$ and $N(I-K)^{\perp}$? Then how does it follow that there exists $u_{k} \in N(I-K)^{\perp}$ solving $u_{k}-Ku_{k} = v_{k}$? 
Thanks for any assistance!   
 A: Ad (1): Gram–Schmidt can readily be applied to any countable collection of non-zero vectors. Just remember how Gram–Schmidt orthonormalisation works: you simply replace $v_{k+1}$ with its orthogonal projection onto $\{v_1,\dotsc,v_k\}^\perp$, and then normalise (if non-zero). Now, suppose that $N(1-K)$ is infinite-dimensional, and that $\{v_k\}$ is a countably infinite basis of $N(1-K)$. Then, in particular, all finite linear combinations of the $\{v_k\}$ are dense in the closed subspace $N(1-K)$, but by Gram–Schmidt, these are exactly the same as the finite linear combinations of the output of Gram–Schmidt orthonormalisation.
Ad (2): By the basic theory of Hilbert spaces, if $U$ is a closed subspace of $H$, then $H = U \oplus U^\perp$, which means, precisely, that for any $x \in H$, you have (unique) $u \in U$, $u^\prime \in U^\perp$, such that $x = u + u^\prime$. Now, apply this to $x_k \in H = N(1-K)^\perp \oplus N(1-K)$ to get $x_k = u_k + u_k^\prime$ for $u_k \in N(1-K)^\perp$ and $u_k^\prime \in N(1-K)$, and observe that
$$
 v_k = (1-K)x_k = (1-K)(u_k + u_k^\prime) = (1-K)u_k + (1-K)u_k^\prime = (1-K)u_k + 0 = (1-K)u_k.
$$
