Since every monomorphism in $Set$ is split, every $F : Set\to Set$ takes monomorphisms to monomorphisms. In particular, your $F(\subseteq)$ is again a monomorphism.

So if you have two coalgebra structures on $U\subseteq A$ that make the square commute, then they must be equal, since you can cancel $F(\subseteq)$ from the left.

The analogous statement is false in categories other than $Set$. For example on $Ab$, consider the functor $-\otimes\mathbb{Z}_2$. The quotient map $\mathbb{Z}\to\mathbb{Z}_2$ makes $\mathbb{Z}$ into a coalgebra. Applying the functor to the inclusion of the even integers $\mathbb{Z}\subseteq\mathbb{Z}$ results in the zero map. Therefore the even integers $\mathbb{Z}$ are a subcoalgebra in two different ways: either via the projection or via the zero map $\mathbb{Z}\to\mathbb{Z}_2$.

Let us say that a monomorphism $U\subseteq A$ is nontrivial if $U$ is nonempty. Since every nontrivial monomorphism in $Set$ is split, every $F : Set\to Set$ takes nontrivial monomorphisms to monomorphisms. In particular, your $F(\subseteq)$ is again a monomorphism in this case.

So if you have two coalgebra structures on $U\subseteq A$ that make the square commute, then they must be equal, since you can cancel $F(\subseteq)$ from the left.

The only case left to treat is the trivial case $U = \emptyset$. But then the coalgebra structure is unique by the initiality.

The uniqueness of the subcoalgebra structure is false in some categories other than $Set$. For example on $Ab$, consider the functor $-\otimes\mathbb{Z}_2$. The quotient map $\mathbb{Z}\to\mathbb{Z}_2$ makes $\mathbb{Z}$ into a coalgebra. Applying the functor to the inclusion of the even integers $\mathbb{Z}\subseteq\mathbb{Z}$ results in the zero map. Therefore the even integers $\mathbb{Z}$ are a subcoalgebra in two different ways: either via the projection or via the zero map $\mathbb{Z}\to\mathbb{Z}_2$.

Since every monomorphism in $Set$ is split, every $F : Set\to Set$ takes monomorphisms to monomorphisms. In particular, your $F(\subseteq)$ is again a monomorphism.

So if you have two coalgebra structures on $U\subseteq A$ that make the square commute, then they must be equal, since you can cancel $F(\subseteq)$ from the left.

Since every monomorphism in $Set$ is split, every $F : Set\to Set$ takes monomorphisms to monomorphisms. In particular, your $F(\subseteq)$ is again a monomorphism.

So if you have two coalgebra structures on $U\subseteq A$ that make the square commute, then they must be equal, since you can cancel $F(\subseteq)$ from the left.

The analogous statement is false in categories other than $Set$. For example on $Ab$, consider the functor $-\otimes\mathbb{Z}_2$. The quotient map $\mathbb{Z}\to\mathbb{Z}_2$ makes $\mathbb{Z}$ into a coalgebra. Applying the functor to the inclusion of the even integers $\mathbb{Z}\subseteq\mathbb{Z}$ results in the zero map. Therefore the even integers $\mathbb{Z}$ are a subcoalgebra in two different ways: either via the projection or via the zero map $\mathbb{Z}\to\mathbb{Z}_2$.