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.
Suppose $T$ is a bounded operator on a Banach space $X$. $\lambda\in\rho(T)$ iff $T-\lambda I$ is a linear bijection. In that case, the inverse $(T-\lambda I)^{-1}$ is automatically continuous by the closed graph theorem.
There are three basic things that can stand in the way of $T-\lambda I$ being invertible.
$T-\lambda I$ is not injective. Equivalently, $Tx=\lambda x$ for some $x\ne 0$, which means that $\lambda$ is an eigenvalue of $T$.
The range of $T-\lambda I$ is not dense in $X$. Equivalently, there is a non-zero bounded linear functional $x^{\star}\in X^{\star}$ such that $x^{\star}((T-\lambda I)y)=0$ for all $y$; this, in turn is equivalent to $(T-\lambda I)^{\star}x^{\star}=0$, which means $\overline{\lambda}$ is an eigenvalue of $T^{\star}$.
$T-\lambda I$ is injective and has dense, non-closed range. In this case, the inverse $T-\lambda I$ cannot be bounded, which gives the existence of a sequence $\{ y_n \}$ of unit vectors in the range of $\mathcal{R}(T-\lambda I)$ such that $\lim_n\|(T-\lambda I)y_n\|=\infty$. After renormalization, you obtain a sequence of unit vectors $\{ x_n \}$ such that $\lim_n\|(T-\lambda I)x_n\|=0$. So $\lambda$ is an approximate eigenvalue of $T$ in this case.
You have an example of (1); there are plenty of examples of eigenvalues.
Case (2) is peculiar to infinite-dimensional spaces. A simple example is the shift operator on $\ell^2(\mathbb{N})$:
$$
U(a_0,a_1,a_2,\cdots) = (0,a_0,a_1,a_2,\cdots)
$$
For $a=(a_0,a_1,a_2,\cdots)$, $\|Ua\|=\|a\|$. So $U$ is injective, and its range is a proper closed subspace of $\ell^2(\mathbb{N})$. You can't do this with a finite shift.
Case (3) is also peculiar to infinite-dimensional spaces. A nice example of this is the multiplication operator $(Mf)(x)=xf(x)$ on $L^{2}[0,1]$. Every $\lambda \in [0,1]$ is an approximate eigenvalue. The simplest case is $\lambda=0$. The functions
$$
f_n = \sqrt{n}\chi_{[0,1/n]},\;\;\; n=1,2,3,\cdots,
$$
are unit vectors because
$$
\int_{0}^{1}|f_n|^2dx = \int_{0}^{1/n}ndx = 1.
$$
And
$$
\|Mf_n\|^2 = \int_{0}^{1}x^{2}|f_n|^2dx = \int_{0}^{1/n}n xdx=
\frac{n}{2}x^{2}|_{0}^{1/n}= \frac{1}{n}\rightarrow 0.
$$
So $0$ is an approximate eigenvalue of $M$. The range is dense because $M=M^{\star}$ means $M^{\star}$ does not have eigenvalue $0$ either. You can see how $\chi_{[\lambda-1/n,\lambda+1/n]}$ is very close to being an eigenfunction of $M$ with eigenvalue $\lambda$, if $\lambda \in [0,1]$, but it just can't quite get there.
Best Answer
In fact, $T$ sends integrable functions to polynomials of at most degree 2. To see that this is the case, note that the integral can be expanded to $$ Tf = \left(\int_0^1 f(t)\,dt\right)s^2 - 2\left(\int_0^1 tf(t)\,dt\right)s + \left(\int_0^1 t^2f(t)\,dt\right). $$ Other than that, your argument is correct. To prove that this kind of operator is generally compact, you could use one of the arguments presented here. In this case, however, it suffices to note that the image of $T$ is finite-dimensional.
From there, we can determine the eigenvectors of $T$ as follows. Consider the restriction of $T$ to the space $\Bbb P_2$ of degree-2 polynomials. Relative to the basis $\mathcal B = (1,t,t^2)$, we find that the matrix of the restriction $T|_{\Bbb P_2}$ is given by $$ [T]_{\mathcal B} = \pmatrix{\frac 13 & \frac 14 & \frac 15\\ -1& -\frac 23 & -\frac 12 \\ 1& \frac 12 & \frac 13}. $$
The eigenvalues of this operator are the non-zero elements of the spectrum of $T$.
You should find that these are equal to $-1/6$ and $(5 \pm 3\sqrt{5})/60$.