Here are the details for the solutions hinted at in the comments. From the general theory about the Fredholm alternative as mentioned in the question, it is sufficient to show that if $f\in C[0,1]$ satisfies
\begin{equation}
f(t) = \int_{0}^{t} K(t,s) f(s) \, ds \tag{1}
\end{equation}
for all $t\in [0,1]$, then $f = 0$.
Solution 1.
Define $M := \sup\{|K(s,t)| : s,t\in [0,1]\}$ and $h\colon [0,1] \to \mathbb{R}$ by $h(t) := \int_{0}^{t} |f(s)| \, ds$. Note that $M < \infty$ because $K$ is continuous on the compact set $[0,1]\times [0,1]$. Also $h$ is continuously differentiable with $h'(t) = |f(t)|$ for every $t\in [0,1]$. Moreover, we also have
\begin{equation}
h'(t) = |f(t)| = \Big| \int_{0}^{t} K(t,s) f(s) \, ds \Big| \leq M \int_{0}^{t} |f(s)| \, ds = M h(t) \tag{2}
\end{equation}
for each $t\in [0,1]$. Define $\ell \colon [0,1] \to \mathbb{R}$ by $\ell (t) = h(t) \exp (-Mt)$. Clearly $\ell (t) \geq 0$ for all $t\in [0,1]$. In addition, note that $\ell$ is continuously differentiable with
\begin{equation}
\ell ' (t) = (h'(t)- M h(t)) \exp (-Mt) \leq 0 \cdot \exp (-Mt) = 0
\end{equation}
for each $t\in [0,1]$, where $(2)$ was used. Hence $\ell$ is decreasing on $[0,1]$ with $\ell (0) = 0$, so it follows that $\ell (t) \leq 0$ for all $t\in [0,1]$. Consequently, we have $\ell (t) = 0$ for all $t\in [0,1]$, which implies $h(t) = 0$ for all $t\in [0,1]$, which implies $|f(t)| = h'(t) = 0$ for all $t\in [0,1]$. Therefore $f = 0$ as desired.
Solution 2.
We show by induction that if $f\in C[0,1]$ satisfies $(1)$, then for every $n\in\mathbb{N}$ we have
\begin{equation}
|f(t)| \leq \frac{(\|K\|_{\infty}t)^{n}}{n!} \|f\|_{\infty} \text{ for all $t\in [0,1]$} , \tag{3}
\end{equation}
where $\|K\|_{\infty} = \sup\{|K(s,t)| : s,t\in [0,1]\}$. Then as
\begin{equation}
\lim_{n\to\infty} \frac{\|K\|_{\infty}^{n}}{n!} = 0
\end{equation}
by this result, it follows from $(3)$ that $f = 0$.
We now show $(3)$. Let $t\in [0,1]$. We have
\begin{equation}
|\Phi (f)(t)| \leq \int_{0}^{t} |K(t,s)| |f(s)| \, ds \leq t \|K\|_{\infty} \|f\|_{\infty} .
\end{equation}
Hence $(3)$ holds with $n=1$. Now suppose $(3)$ holds for some $n\in\mathbb{N}$. Let $t\in [0,1]$. We have
\begin{align*}
|f(t)| = \Big| \int_{0}^{t} K(t,s) f(s) \, ds \Big| &\leq \int_{0}^{t} |K(t,s)| |f(s)| \, ds \\
&\leq \frac{\|K\|_{\infty}^{n}}{n!} \int_{0}^{t} \|K\|_{\infty} s^{n} \, ds = \frac{(\|K\|_{\infty} t)^{n+1} }{(n+1)!} .
\end{align*}
Hence $(3)$ holds for $n+1$. This completes the induction.
