You can prove more. Let $F(u,v)$ be any $1$-Lipschitz function on $[0,1]^2$ such that $F(u,v)<\min(u,v)$ inside the square. Then there exists a copula $D(u,v)$ such that
$$
F(u,v)\le D(u,v)<\min(u,v)
$$
everywhere inside the square.
The second inequality will be immediate if we just construct the corresponding joint distribution measure $\mu$ on $[0,1]^2$ with full support. Indeed, then
$$
D(u,v)=\mu([0,u]\times [0,v])=\mu([0,u]\times [0,1])-\mu([0,u]\times(u,1])
\\
=u-\mu([0,u]\times(u,1])<u
$$
and similarly for $v$.
It remains to make the following two observations.
Observation 1: There is a copula $D_0(u,v)>F(u,v)$ inside the unit square such that the associated $\mu$ has density $p_0(u,v)\ge q_0>0$ separated away from $0$ in some open neighborhood $\Omega$ of $\{(x,x):0<x<1\}$. To see it, just take a sufficiently fine countable partition of $(0,1)$ into disjoint intervals $I_j$, consider the density $p=\sum_j|I_j|^{-1}\chi_{I_j\times I_j}$ and mix two such distributions to take care of the corners. "Sufficiently fine" just means that the length of each $I_j$ is much less than $u-F(u,u)$ for $u\in I_j$. Now take any sequence $q_0>q_1>\dots \ge q_0/2$ and
Observation 2: Suppose we have a copula $D_n$ for which $D_n(u,v)>F(u,v)$ inside the unit square and the density $p_n$ of $\mu_n$ is at least $q_n$ in $\Omega$. Let $(u_n,v_n)\in(0,1)^2$ be any point outside the diagonal. Then, for every $\delta_n\ge 0$, there exists a copula $D_{n+1}$ with $D_{n+1}>F$ inside the unit square, such that $p_{n+1}\ge p_n$ outside $\Omega$, $p_{n+1}\ge q_{n+1}$ in $\Omega$, and $(u_n,v_n)$ is in the support of $\mu_{n+1}$. To do it, just choose two very short intervals $U,V$ of equal length containing $u_n$ and $v_n$ respectively and set
$$
p_{n+1}=p_n+t(-\chi_{U\times U}-\chi_{V\times V}+\chi_{V\times U}+\chi_{U\times V})
$$
with very small $t$. Notice that this changes $D_n(u,v)$ only on a compact subset of the open unit square (say, $I\times I$ where $I$ is a closed interval such that $U\cup V\subset I\subset (0,1)$) and there the difference $D_n-F$ is separated from $0$, so this surgery leaves it positive for small $t>0$.
Now just take any sequence $(u_n,v_n)\in (0,1)^2\setminus{\rm diag}$ dense in the square and run this recursion. Then take either the weak limit of $\mu_n$, or the $L^1$-limit of $p_n$, whichever you are more comfortable with, to get $D$.
