Let $t_{0}\in[0,1]$ be given. Let $C^{1}[0,1]$ be the commutative Banach algebra consisting of continuously differentiable complex valued functions on $[0,1]$, endowed with the norm $$\|f\|:=\|f\|_{\infty}+\|f'\|_{\infty}.$$ I have shown that $$I(t_{0}):=\{f\in C^{1}[0,1]:f(t_{0})=f'(t_{0})=0\}$$ is an (algebra-)ideal of $C^{1}[0,1]$ and that $$A:=C^{1}[0,1]/I(t_{0})$$ is a 2-dimensional Banach algebra. How do I prove that the radical $\text{Rad}(A)$ is $1$-dimensional. Recall that $\text{Rad}(A)$ is the intersection of all maximal ideals in $A$, or equivalently, the intersection of all kernels of (unit preserving) algebra homomorphisms $h\colon A\to\mathbb{C}$.
I tried to construct all possible homomorphisms. I think (but I'm not entirely sure) that $1+I(t_{0})$ and $x+I(t_{0})$ form a basis of $A$ (where $x:t\mapsto t$), so any homomorphism $h$ is determined by the image of $x+I(t_{0})$ (since $h(1+I(t_{0}))$ is always $1$). But then I find that the radical is $\{I(t_{0})\}$, which is zero dimensional.
Hints: One can show that there is only one (unit preserving) algebra homomorphism $h:A\to \mathbb C$.
There are probably multiple ways to show this. As you suggested, $1+I(t_0)$ and $x+I(t_0)$ form indeed a basis of $A$. One can show this by using the definition of linear independence (since you already know that $A$ is 2-dimensional this should be enough). Then you can calculate the coefficients for $x^2+I(t_0)$ of that basis. Note that $$ h(x^2+I(t_0))=h(x+I(t_0))^2 $$ has to be true because $h$ is an algebra homomorphism. Using this information it turns out there is only one possibility for $h$.
It turns out that this $h$ is given via $$ h(f + I(t_0)) = f(t_0) $$ where $f\in C^1[0,1]$ is a function. It can then be shown that $h:A\to \mathbb C$ is well-defined (because $g(t_0)=0$ if $g\in I(t_0)$) and an algebea homomorphism (this follows because the pointwise evaluation preserves multiplication and addition).
Finally, $\operatorname{Rad}(A)$ can be calculated.