Example to prove that $ C^1[0,1] $ is not a Banach space for the uniform norm?

The example you gave converges uniformly to the zero function, which is continuously differentiable.

Every continuous function on $[0,1]$ is a uniform limit of polynomial functions (by the Weierstrass approximation theorem), and polynomial functions are continuously differentiable.

For an explicit example, you could also consider the sequence $f_n(x)=\left|x-\frac12\right|^{(n+1)/n}$.

---

I might not fully understand the last question. Those are not the same, because elements of $C^1[0,1]$ do not generally have to vanish at $0$. But you can use similar examples. E.g., you could still think about the Weierstrass approximation theorem for dramatic counterexamples, or you could modify the example above by taking, say, $g_n(x)=f_n(x)-f_n(0)$.

The ${\cal C}^1[0,1]$ functions are a dense subset of $\cal{C}[0,1]$ in the sup norm. However the inclusion is proper so the $\cal{C}^1$ functions are not a complete subspace of ${\cal C}[0,1]$.

The continuous functions on $[0,1]$ that vanish at 0 are a Banach space, they are the kernel of the continuous map $f\mapsto f(0)$. They form a closed subspace. But those in ${\cal C}^1[0,1]$ are not.