For $V$ finite-dimensional, the algebra $\hom(V, V) \cong V^\ast \otimes V$ is naturally self-dual, and this duality may be used to transfer the algebra structure on $\hom(V, V)$ to a coalgebra structure on $\hom(V, V)$, and vice-versa. Notice that the duality functor $\text{Vect}^{op} \to \text{Vect}$ induces a functor from coalgebras to algebras.

Thus a coalgebra map $M_n(\mathbb{C}) \to M_{n+1}(\mathbb{C})$ would induce a dual algebra map $M_{n+1}(\mathbb{C}) \to M_n(\mathbb{C})$. But even when $n=1$ there is no such algebra map.

I don't know whether the space of compact operators carries a (natural) coalgebra structure.

For $V$ finite-dimensional, the algebra $\hom(V, V) \cong V^\ast \otimes V$ is naturally self-dual, and this duality may be used to transfer an algebra structure on $\hom(V, V)$ to a coalgebra structure on $\hom(V, V)$, and vice-versa. (Notice that the duality functor $\text{Vect}^{op} \to \text{Vect}$ on finite-dimensional spaces induces a functor from coalgebras to algebras, and vice-versa.)

The map $M_n(\mathbb{C}) \to M_{n+1}(\mathbb{C})$ mapping $A \mapsto A \oplus 0$ is dual to the map $M_{n+1}(\mathbb{C}) \to M_n(\mathbb{C})$ that takes an $(n+1) \times (n+1)$ matrix to the $n \times n$ matrix made from the first $n$ rows and columns. If we choose the algebra structure on the spaces $M_n(\mathbb{C})$ to be not the usual matrix multiplication but the one given by entrywise multiplication, then these are algebra maps and they dualize to coalgebra maps. This gives an affirmative answer to the question. However, this is clearly highly dependent on basis and doesn't strike me as terribly interesting.

The forgetful functor $\text{Coalg} \to \text{Vect}$ creates colimits, and so the colimit of a chain of coalgebra maps $M_n(\mathbb{C}) \to M_{n+1}(\mathbb{C})$ is created from the colimit in $\text{Vect}$. This answers the second question affirmatively.