Answer: no. Suppose that the segment $[0, 1]$ can be represented as the union $A \cup B$ of two disjoint sets $A$ and $B$ such that $f(A) = B$. Let us immediately note that the function $f$ is strictly increasing on the interval $[0, 1]$ and is a bijection from $[0, 1]$ to $f([0, 1])$.
Let us prove that $f$ has no fixed points on the interval $[0, 1]$. Indeed, let $f(s) = s$ for some $s \in [0, 1]$. If $s \in A$, then $f(s) = s \notin B$, and if $s \in B$, then there is no number from $A$ that is sent to $s$, because $f$ is a bijection. Since both cases are impossible and $[0, 1] = A \cup B$, we arrive at a contradiction, that is, there cannot be fixed points. Due to the continuity of the function, this means that $f(x) > x$ for all $x \in [0, 1]$ or $f(x) < x$ for all $x \in [0, 1]$ depending on the sign of $c$.
**Case $c \ge 0$.** We have $x < f(x)$ for all $x \in [0, 1]$. Since $f(0) = c > 0$ and the function is increasing on the interval $[0, 1]$, then $f(x) \ge c$ for any $x \in [0, 1]$. Therefore $[0, c[ \subseteq A$, whence $[c, f(c)][ \subseteq B$, and $c < f(c)$. Let us now consider the interval $[f(c), f(f(c))[$. Since all points of this interval are images of points from $B$ and not from $A$ (after all, $f$ is a bijection), the intersection of $[f(c), f(f(c))[$ with $B$ is empty. We know that $f(c) \le 1$ and $f(c) < f(f(c))$. Note that $1 \notin [f(c), f(f(c))[$, because otherwise we would get that $1 \notin B$, whence $1 \in A$ and $1 < f(1) \in B$, which is impossible. Consequently, $[f(c), f(f(c))[$ $\subseteq A$ and, continuing the reasoning in this way, we obtain that the sequence
$0, f(0), f^2(0), f^3(0), \dots$
increases and lies entirely in $[0, 1]$, that is, it is bounded above. This means that it has a limit $\ell \in [0, 1]$. From the continuity of $f$ we obtain that
$$\ell = \lim_{n \to \infty} f^n(0) = f\left(\lim_{n \to \infty} f^{n-1}(0)\right) = f(\ell),$$
that is, $\ell$ is a fixed point. Contradiction.
**Case $c < 0$.** We have $f(x) < x$ for all $x \in [0, 1]$. In particular, $f(1) < 1$. The function is increasing on the interval $[0, 1]$, and so $f(x) \le f(1)$ for any $x \in [0, 1]$. Therefore $]f(1), 1] \subseteq A$, whence $]f(f(1)), f(1)] \subseteq B$, and $f(f(1)) < f(1)$. Similar to the previous case, we show that $]f^3(1), f^2(1))[$ $\subseteq A$ and that the sequence
$$1, f(1), f^2(1), f^3(1), \dots$$
converges to some number $\ell \in [0, 1]$, which is a fixed point for $f$. Contradiction.