**Step 1: Show $0 \in \sigma(K)$.**

Suppose for contradiction that $0 \in \rho(K)$, i.e., $K$ is bijective with bounded inverse $K^{-1} \in \mathcal{L}(X)$. Then the identity operator satisfies \begin{align*} I = K^{-1} \circ K. \end{align*} Since $K$ is compact and $K^{-1}$ is bounded, the composition $K^{-1} \circ K$ is compact. But the identity on an infinite-dimensional Banach space is not compact (the sequence of unit vectors in any infinite orthonormal set has no convergent subsequence). This is a contradiction, so $0 \in \sigma(K)$.

**Step 2: Every $\lambda \in \sigma(K) \setminus \{0\}$ is an eigenvalue.**

Let $\lambda \in \mathbb{C}$ with $\lambda \ne 0$ and $\lambda \in \sigma(K)$. Write $K - \lambda I = -\lambda(I - \lambda^{-1} K)$. Set $K_\lambda := \lambda^{-1} K$, which is compact since $K$ is compact and $\lambda^{-1}$ is a nonzero scalar. Then $\lambda \in \sigma(K)$ means $I - K_\lambda$ is not bijective.

By the Fredholm alternative (applied to the compact operator $K_\lambda$), if $N(I - K_\lambda) = \{0\}$ then $I - K_\lambda$ is bijective. Since it is not, we must have $N(I - K_\lambda) \ne \{0\}$, i.e., there exists $u \ne 0$ with $(I - K_\lambda)u = 0$, which gives $u = \lambda^{-1}Ku$, i.e., $Ku = \lambda u$. Hence $\lambda$ is an eigenvalue: $\lambda \in \sigma_p(K)$.

**Step 3: Every eigenspace at a nonzero eigenvalue is finite-dimensional.**

Let $\lambda \ne 0$ be an eigenvalue of $K$ and let $E_\lambda := N(K - \lambda I)$ be the corresponding eigenspace. The closed unit ball $B_\lambda := \{u \in E_\lambda : \|u\|_X \le 1\}$ satisfies $K(B_\lambda) = \lambda B_\lambda$ (since $Ku = \lambda u$ for $u \in E_\lambda$). Hence $B_\lambda = \lambda^{-1} K(B_\lambda)$, which is precompact since $K$ is compact and $\lambda^{-1}$ is bounded. A normed space with precompact unit ball is finite-dimensional, so $\dim E_\lambda < \infty$.

**Step 4: The set $\sigma(K) \setminus \{0\}$ has no accumulation point other than $0$.**

Suppose for contradiction that there exists a sequence of distinct eigenvalues $\{\lambda_k\}_{k=1}^\infty \subseteq \sigma_p(K) \setminus \{0\}$ with $\lambda_k \to \mu \ne 0$. For each $k \in \mathbb{N}$, let $u_k \in X$ be a unit eigenvector with $Ku_k = \lambda_k u_k$. Define the finite-dimensional subspaces \begin{align*} E_k := \mathrm{span}\{u_1, u_2, \ldots, u_k\}. \end{align*} Eigenvectors corresponding to distinct eigenvalues of a linear operator are linearly independent, so $\dim E_k = k$ and $E_k \subsetneq E_{k+1}$ for every $k \in \mathbb{N}$.

By the Riesz lemma, for each $k \ge 2$ there exists $v_k \in E_k$ with $\|v_k\|_X = 1$ and $\mathrm{dist}(v_k, E_{k-1}) \ge \frac{1}{2}$. Write $v_k = \sum_{j=1}^k \alpha_j^{(k)} u_j$ for scalars $\alpha_j^{(k)} \in \mathbb{C}$.

For $k > m$, compute \begin{align*} Kv_k - Kv_m &= \lambda_k v_k - Kv_m. \end{align*} The term $\lambda_k v_k - Kv_m$ can be rearranged as $\lambda_k v_k - (Kv_m - \lambda_k v_m) - \lambda_k v_m$. The expression $Kv_m - \lambda_k v_m = (K - \lambda_k I)v_m$ lies in $E_{k-1}$ because $v_m \in E_m \subseteq E_{k-1}$ and $(K - \lambda_k I)$ maps $E_{k-1}$ into itself (since $Ku_j = \lambda_j u_j \in E_{k-1}$ for $j < k$). Similarly $\lambda_k v_m \in E_{k-1}$. Therefore $Kv_k - Kv_m - \lambda_k v_k \in E_{k-1}$, which gives \begin{align*} \frac{1}{\lambda_k}(Kv_k - Kv_m) = v_k - w \quad \text{for some } w \in E_{k-1}. \end{align*} Hence \begin{align*} \left\|\frac{Kv_k}{\lambda_k} - \frac{Kv_m}{\lambda_m}\right\|_X \ge \left\|\frac{Kv_k}{\lambda_k} - \frac{Kv_m}{\lambda_m}\right\|_X. \end{align*} More directly, since $\mathrm{dist}(v_k, E_{k-1}) \ge \frac{1}{2}$ and $\frac{Kv_m}{\lambda_k} + (w - v_k) + \frac{Kv_m}{\lambda_k} - \frac{Kv_m}{\lambda_m} \in E_{k-1}$: \begin{align*} \left\|\frac{Kv_k}{\lambda_k} - \frac{Kv_m}{\lambda_m}\right\|_X \ge \mathrm{dist}(v_k, E_{k-1}) \ge \frac{1}{2}. \end{align*} Since $\lambda_k \to \mu \ne 0$, the sequence $\{\lambda_k^{-1} K v_k\}_{k=1}^\infty = \{v_k - \text{(correction in } E_{k-1})\}$ is bounded, yet no subsequence is Cauchy (distances between terms are at least $\frac{1}{2}$). This contradicts compactness of $K$ applied to the bounded sequence $\{v_k\}$. Therefore $\sigma(K) \setminus \{0\}$ has no accumulation point other than $0$.

**Step 5: Conclude the structure of $\sigma(K)$.**

From Steps 2 and 4: every element of $\sigma(K) \setminus \{0\}$ is an eigenvalue (Step 2), each eigenvalue has finite-dimensional eigenspace (Step 3), and the eigenvalues cannot accumulate at any nonzero point (Step 4). A subset of $\mathbb{C}$ with no accumulation point other than $0$ is either finite or countably infinite, in the latter case forming a sequence converging to $0$.