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.
The resolvent $(T-\lambda I)^{-1}$ is obtained by solving for $g$ in the following:
$$
g = (T-\lambda I)^{-1}f \\
(T-\lambda I)g= f \\
\int_0^x g(t)dt-\lambda g(x)=f(x) \\
\int_0^xg(t)dt-\lambda\frac{d}{dx}\int_0^x g(t)dt=f(x) \\
\frac{d}{dx}\int_0^xg(t)dt-\frac{1}{\lambda}\int_0^x g(t)dt=-\frac{1}{\lambda}f(x) \\
\frac{d}{dx}\left(e^{-x/\lambda}\int_0^xg(t)dt\right)=-\frac{1}{\lambda}e^{-x/\lambda}f(x) \\
e^{-x/\lambda}\int_0^x g(t)dt = -\frac{1}{\lambda}\int_0^x e^{-u/\lambda} f(u)du \\
\int_0^xg(t)dt=-\frac{1}{\lambda}\int_0^x e^{(x-u)/\lambda}f(u)du \\
g(x)=-\frac{1}{\lambda}f(x)-\frac{1}{\lambda^2}\int_0^x e^{(x-u)/\lambda}f(u)du
$$
So,
$$
(T-\lambda I)^{-1}f = -\frac{1}{\lambda}f(x)-\frac{1}{\lambda^2}\int_0^x e^{(x-u)/\lambda}f(u)du
$$
and the spectrum is $\sigma(T)=\{0\}$.
Best Answer
The kernel of $V$ is $K(x,t) = \mathsf 1_{(0,x)}(t)$ and $$ \int_{[0,1]^2} |K(x,t)|^2\ \mathsf d(x\times t) = \int_0^1\int_0^x\ \mathsf dt \ \mathsf dx = \frac 12<\infty, $$ so $V\in L^2([0,1]^2)$. It follows that $T$ is a Hilbert-Schmidt operator and therefore is compact. It follows then from the Fredholm alternative that if $\lambda\in\sigma(V)$, $\lambda\ne0$, that $\lambda$ is an eigenvalue of $V$. That is, $Vf(x)=\lambda f(x)$ for all $x\in[0,1]$. If $Vf=\lambda f$, then as $Vf\in C^0([0,1])$ (we can prove that $V$ is Hölder continuous using the Hölder inequality), we must have $f\in C^0([0,1])$. By the fundamental theorem of calculus, $Vf\in C^1([0,1])$, and so $f\in C^1([0,1])$. Differentiating both sides of the eigenvalue equation yields $f(x)= \lambda f'(x)$, and hence $f(x) = Ce^{\frac1\lambda x}$ for some $C\in\mathbb C$. Since $Vf(0) = 0$, we have $f(0)=0$, and therefore $C=0$. It follows that the spectral radius of $V$ is zero.