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.
This just a way to write the spectral theorem:
https://en.wikipedia.org/wiki/Compact_operator_on_Hilbert_space#Spectral_theorem
Since $0$ is an isolated element,the orthogonal space of $Ker T$ cannot be infinite dimensional, since from the spectral theorem, in this situation the orthogonal space of $Ker T$ would have an orthogonal basis $e_i$ with $T(e_i)=c_ie_i, lim_nc_n=0$ this is in contradiction with the fact that $0$ is isolated, thus the spectral theorem implies that the orthogonal of $Ker T$ is finite dimensional and $Ker T$ infinite dimensional.
Best Answer
Let $M$ be the kernel of $K$. Then $S:H/M \to K(H)$ defined by $S(x+M)=Kx$ is a well defined bounded operator which is also bijective. By Open Mapping Theorem its inverse is also continuous. Since $K$ is compact this implies that the closed unit ball of $K(x)$ is compact and hence $K(X)$ is finite dimensional.
Some details for the last part: suppose $(y_n)$ is a bounded sequence in $K(X)$. Let $x_n=S^{-1}y_n$. Then $\|x_n\| \leq \|S^{-1}\| \|y_n\|$ so $(x_n)$ is bounded. By the definition of the norm in $H/M$ we can find a sequence $(z_n)$ in $M$ such that $(x_n+z_n)$ is bounded. Since $K$ is compact $K(x_n+z_n)$ has a convergent subsequence. But $Kz_n=0$ so $(K(x_n))$ has a convergent subsequence. By definition of $x_n$ we have $y_n=S(x_n)=Kx_n$. Hence $(y_n)$ has a convergent subsequence. Thus every bounded sequence in $K(H)$ has a convergent subsequence. This implies that $K(H)$ is finite dimensional.