Consider the map (as defined by kahen in a comment)
$$ T_h f(x) = h(x) + \int_0^1 K(x+y) f(y) dy $$
which is a linear map from $C([0,1])$ to itself. Then
$$ T_h f(x) - T_h g(x) = \int_0^1 K(x+y) (f(y) - g(y)) dy $$
$$ | T_h f(x) - T_h g(x) | \leq \int_0^1 K(x+y) |f(y) - g(y)| dy$$
where we used that $K$ is positive. Now, take $|f-g|$ in $\sup$ over $[0,1]$, and use that $K$ is strictly decreasing, so $\int_0^1 K(x+y) dy \leq \int_0^1 K(y) dy < 1$, you have that for any $h$, $T_h$ is a contraction mapping.
Now just apply the Banach fixed-point theorem and you are done.
There seems to be some confusion about whether $K$ is taken to be strictly decreasing. Note that in the crucial aspect of the proof above is that $\int_0^1 K(x+y) \leq 1-\epsilon $ for some positive $\epsilon$. That $K$ is strictly decreasing is sufficient, but not necessary.
On the other hand, TonyK indicated in his comment that in the case that $K\equiv 1$ and $h = 0$ there are infinitely many solutions (take $u$ to be any constant function). Analogous to the case of linear algebra where a change of parameter can change a linear system from having infinitely many solutions to no solutions, we see the same phenomenon here. Let $K\equiv 1$ and assume $h$ is a function with integral $\int_0^1 h(x) dx \neq 0$. Then the given integral equation admits no solutions.
Since $K \equiv 1$, the equation reduces to
$$ u(x) = h(x) + \int_0^1 u(y) dy $$
Integrate both sides in $x$, you get
$$ \int_0^1 u(x) dx = \int_0^1 h(x) dx + \int_0^1 dx \int_0^1 u(y) dy $$
so we can subtract the left hand side from the right hand side and get
$$ 0 = \int_0^1 h(x) dx \neq 0 $$
which is absurd. In fact, you have the following claim: if $K\equiv 0$, the equation either has no solutions (when $h$ has non-zero average) or infinitely many solutions (when $h$ has zero average) differing from each other by constant shifts.
So by the principle of "homework exercises should be solvable", you can probably assume that whatever the source of the problem, the phrase "decreasing" is intended to mean "strictly decreasing".