[Solution] The minimum value of $c$ is 2.
In fact, consider the function
$$
f(x)=\left\{\begin{array}{l}
0, \quad 0 \leqslant x \leqslant \frac{1}{2} ; \\
1, \quad \frac{1}{2}<x \leqslant 1.
\end{array}\right.
$$

If $x > \frac{1}{2}$, then $f(x) \leqslant 1 < 2x$.
If $0 < x \leqslant \frac{1}{2}$, then we can choose $n$ such that
$$
\frac{1}{2^{n+1}} < x \leqslant \frac{1}{2^{n}},
$$

Thus, we have
$$
\begin{array}{c}
2^{n} f(x) \leqslant f\left(2^{n} x\right) \leqslant 1 < 2 \cdot 2^{n} x, \\
f(x) < 2 x.
\end{array}
$$

In summary, $f(x) \leqslant 2 x$ holds for all $x \in [0,1]$.
$6 \cdot 91$ Consider the function $f(x)$ on $[0,1]$ satisfying the conditions:
(1) $f(x) \geqslant 0$, for any $x \in[0,1]$;
(2) $f(1)=1$;
(3) $f(x)+f(y) \leqslant f(x+y), x, y, x+y \in[0,1]$

Try to find the smallest constant $c$, such that for any function $f(x)$ satisfying (1) - (3), we have $f(x) \leqslant c x, x \in[0,1]$.