Spectrum of integral operator $A$: $(A f)t = \int_0^t f$ by definition of $Sp$

Why is $A$ not invertible?

Note that for any element $g$ of the image of $A$ (i.e. $g = Af$ for some $f$), we have $g(0) = 0$. From this, it is clear that $A$ is not onto and therefore not invertible.

Why is $A - \lambda I$ invertible for all $\lambda \neq 0$?

We want to find an inverse for $(A - \lambda I)$. Correspondingly, we want a solution to $(A - \lambda I) f = g$ for every $g \in C([0,1],\Bbb C)$. First, consider the case where $g$ is differentiable. We have $$ - \lambda f(t) + \int_0^t f(x)\,dx = g(t) \iff - \lambda f'(t) + f(t) = g'(t) \quad \text{and} \quad f(0) = -(1/\lambda) g(0). $$ We can solve this differential equation as follows: $$ - \lambda f'(t) + f(t) = g'(t)\\ [e^{-\lambda t} f(t)]' = e^{-\lambda t}g'(t)\\ e^{-\lambda t}f(t) = e^{-\lambda \cdot 0}f(0) + \int_0^x e^{-\lambda x}g'(x) \,dx = - \frac 1{\lambda}g(0) + \int_0^x e^{-\lambda x}g'(x) \,dx\\ f(t) = -\frac 1{\lambda}e^{\lambda t} g(0) + e^{\lambda t} \int_0^x e^{-\lambda x}g'(x) \,dx. $$ So, let $B$ be the map defined by the Riemann Stieltjes integral $$ (Bg)(t) = -\frac 1{\lambda}e^{\lambda t} g(0) + e^{\lambda t} \int_{x = 0}^{t} e^{-\lambda x}dg(x). $$ Note that $B$ is a bounded (i.e. continuous) linear map. With the help of the above work, we can see that for all differentiable $g$, we have $(A - \lambda I)Bg = g$ and $B(A - \lambda I)g = g$. Because the differentiable functions form a dense subspace of $C([0,1],\Bbb C)$, we can conclude by continuity that $(A - \lambda I)Bg = g$ and $B(A - \lambda I)g = g$ hold for all elements $g$ of $C([0,1],\Bbb C)$.

So, $A - \lambda I$ is invertible with inverse $B$.