= Solution
We first prove the needed <Fredholm alternative for a compact operator> without assuming self-adjointness. If $C$ is compact, the <finite-rank approximation theorem for compact operators on a Hilbert space> follows directly by covering the compact closure of $C$'s unit-ball image by finitely many balls of radius $\varepsilon$, spanning their centers by a finite-dimensional space $E$, and using the <orthogonal projection> $P_E$. Then $\|C-P_EC\|\leq\varepsilon$.
Choose a finite-rank $F$ with $\|C-F\|<1$, put $E_0=C-F$, and factor
$$
I-C=(I-E_0)(I-G),\qquad G=(I-E_0)^{-1}F.
$$
The first factor is invertible by the <Neumann series>, and $G$ has finite rank. With $V=\operatorname{ran}G$ and $H=V\oplus V^\perp$, the second factor has the triangular form
$$
I-G=
\begin{pmatrix}I_V-G|_V&-G|_{V^\perp}\\0&I_{V^\perp}\end{pmatrix}.
$$
It is injective precisely when the finite-dimensional block $I_V-G|_V$ is injective, and onto precisely when that block is onto. These properties are equivalent in finite dimension. This proves that $I-C$ is injective if and only if it is onto; in that case its inverse is bounded.
For $\lambda\ne0$, apply this result to $C=K/\lambda$. If $\lambda$ lies in the <spectrum of a bounded operator> $K$, then $K-\lambda I$ is not invertible, hence is not injective. Therefore \b[every nonzero spectral point of $K$ is an eigenvalue]. On its <eigenspace> $E_\lambda$, the operator $K$ equals $\lambda I$. Compactness of $K$ would make the closed unit ball of $E_\lambda$ compact after dividing by $\lambda$; part (a) forces $\dim E_\lambda<\infty$.
It remains to rule out infinitely many distinct <eigenvalues> away from zero. Suppose $|\lambda_n|\geq\varepsilon>0$ are distinct, with respective eigenvectors $v_n$. Distinct eigenvalues give linearly independent eigenvectors. Let $V_n=\operatorname{span}(v_1,\ldots,v_n)$ and choose a unit vector $h_n\in V_n\cap V_{n-1}^\perp$. Since $V_n$ is invariant and $(K-\lambda_nI)V_n\subset V_{n-1}$,
$$
Kh_n=\lambda_nh_n+w_n,\qquad w_n\in V_{n-1}.
$$
For $m<n$, $Kh_m\in V_{n-1}$, and <orthogonality> gives $\|Kh_n-Kh_m\|\geq|\lambda_n|\geq\varepsilon$. This contradicts compactness. Thus for each $\varepsilon>0$ there are only finitely many distinct spectral points with modulus at least $\varepsilon$.
Taking $\varepsilon=1/j$ shows that the nonzero <spectrum> is countable; if it is infinite, its distinct elements can be enumerated as a sequence tending to zero. Since the <spectrum> is closed, zero then also belongs to it. In an infinite-dimensional space zero belongs to the spectrum even if there are only finitely many nonzero spectral points: an invertible compact $K$ would make $I=K^{-1}K$ compact, contradicting part (a). \b[The only possible accumulation point is zero, and every nonzero eigenspace is finite-dimensional.] Zero need not be an eigenvalue. The argument uses invariant spans, not an orthogonal eigenbasis, which is unavailable for a general non-self-adjoint compact operator.
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