For compactness you can invoke a general theorem: Hilbert-Schmidt integral operators are compact. Indeed, by changing the order of integration your operator can be written as $Cf(x)=\int_0^1 k(x,y) f(y)\,dy$ with $k(x,y)=-\min(x,y)$. Thus, $C$ is compact and self-adjoint.
The idea to use the fact $(Cf)''=-f$ is sound. Regularity should be briefly discussed: we see that $Cf$ is always continuous; hence, $Cf=\lambda f$ implies that $f$ is continuous; from here we get that $Cf$ is in class $C^1$, etc... the conclusion is that eigenfunctions are $C^\infty$ smooth.
It is important to note the fact that not every every function $g$ with $\lambda g''=-g$ (i.e., a candidate for eigenfunction) is in the range of $C$. Indeed, $Cf(0)=0$ and $(Cf)'(1)=0$ by the definition of $Cf$. These boundary conditions will limit the set of eigenfunctions to a discrete (but still infinite) family.
Since $K$ is compact, $0\in\sigma(K)$ (because $K$ is not invertible). Because $K$ is compact, any nonzero element of its spectrum has to be an eigenvalue.
Now suppose that $Ku=\lambda u$ for some $\lambda\ne0$. This, with your specific $k$, looks like
$$
\lambda u(x)=\int_0^x y\,u(y)\,dy-x\,\int_1^x u(y)\,dy.
$$
Since $u$ is integrable (otherwise $Ku$ makes no sense), the right-hand-side is continuous; so $u$ is continuous. But then the right-hand-side is differentiable, so $u$ is differentiable. Note also that $u(0)=0$, again because the right-hand-side is $0$ at $x=0$.
Now, if we differentiate,
$$
\lambda u'(x)=x\,u(x)-\int_1^xu(y)\,dy-x\,u(x)=-\int_1^xu(y)\,dy.
$$
Reasoning as before, we deduce that $u'(1)=0$, and that $u'$ is differentiable. Taking derivatives again,
$$
\lambda u''(x)=-u(x).
$$
The case $\lambda=0$ gives $u=0$, so $0$ is not an eigenvalue (it belongs to the spectrum, though). When $\lambda\ne0$,
this is an easy second order boundary value problem:
$$
u''+\frac1\lambda\,u=0,\ \ u(0)=0, \ u'(1)=0.
$$
The general solution is, if we write $r=1/\sqrt\lambda$,
$$
u(x)=\alpha\cos rx+\beta\sin rx.
$$
The initial conditions force $\alpha=0$, $\cos r=0$. So $r=\frac{2k+1}2\pi$, $k\in\mathbb Z$, that is
$$
\frac1{\sqrt\lambda}=\frac{2k+1}2\pi,
$$
so the eigenvalues are given by
$$
\lambda_n=\frac4{(2n+1)^2\pi^2}, \ n\in\mathbb N\cup\{0\},
$$
with corresponding eigenfunctions
$$
u_n(x)=\sin\frac{(2n+1)\pi}2\,x
$$
Best Answer
Observe \begin{align} Tf(x)=\int^\pi_0 \sin(x-y) f(y)\ dy = \left(\int^\pi_0 \cos(y)f(y)\ dy\right) \sin(x)-\left(\int^\pi_0 \sin(y)f(y)\ dy\right)\cos(x) \end{align} which means $\mathcal{R}(T) = \operatorname{span}\{\cos(x), \sin(x)\}$ and $\dim\mathcal{R}(T)=2$. Next, consider $f(x) = A\cos(x)+ B\sin(x)$, then we see that \begin{align} Tf(x) = \frac{\pi A}{2}\sin(x)-\frac{\pi B}{2}\cos(x) = \lambda (A\cos(x)+ B\sin(x)) \end{align} iff $\lambda A = -\frac{\pi B}{2}$ and $\lambda B= \frac{\pi A}{2}$. Solving for $A$ and $B$ yields \begin{align} \lambda^2 +\frac{\pi^2}{4} =0 \ \ \implies \ \ \lambda = \pm \frac{\pi}{2}i \end{align} which means $f_1(x) = \cos(x)-i\sin(x)$ and $f_2(x) = \cos(x)+i\sin(x)$ are both eigenfunctions of $T$.
Remark: Note that if $L^2(0, \pi)$ is viewed as a real vector space, then $T$ has no real eigenvalue other than $0$.