If $ k=0 $ by the Ascoli - Arzelá theorem the answer is yes, but i don't know how to proceed in the general case ($ k> 0, k \in \mathbb{N} $). I tried to build a counter example using Riesz lemma but the fact that the image of the succession "lose" an amount of its norm doesn't allow me to control the distance between the images of the single elements and find an absurd. Are there some theorem or classical counter example which I am missing?
Obviously $ C^k $ is endowned with the usual norm given by the sum of the $ C^0 [0, 1]$ norm of every $ n-$derivate for every $0 \leq n\leq k $
We can use Arzelà-Ascoli for each derivative. More precisely, let $(g_n)_{n\geqslant 1}$ be a bounded sequence in $C^{k+1}$. The sequence $(g_n^{(k)})_{n\geqslant 1}$ satisfies the assumptions of Arzelà-Ascoli theorem, hence we can extract a subsequence $(g_{i}^{(k)})_{i\in A_k}$ converging to $f_k$ uniformly on $[0,1]$. Here $A_k$ is an infinite subset of the set of positive integers. Consider now $(g_i^{(k-1)})_{i\in A_k}$ and take $A_{k-1}\subset A_k$ infinite such that $(g_i^{(k-1)})_{i\in A_{k-1}}$ is convergent to some $f_{k-1}$.
At the end, we get $A_0\subset \mathbb N$ infinite such that for each $j\leqslant k$, the sequence $(g_i^{(j)})_{i\in A_0}$ converges uniformly to $f_j$. It remains now to prove that $(f_0)^{(j)}=f_j$ for each $j$.