Well, $D$ is a subcoalgebra of $C$ iff $\Delta(D)\subseteq D\otimes D$, where $\Delta:C\to C\otimes C$ is the comultiplication of $C$.
So... you have to check that condition.
For example, suppose $X$ is a set and $kX$ is the grouplike coalgebra on $X$, so that $kX$ is the vector space which has $X$ as a basis, $\Delta:kX\to kX\otimes kX$ is the unique linear map such that $\Delta(x)=x\otimes x$ for all $x\in X$, and $\epsilon:kX\to k$ is the unique linear map such that $\epsilon(x)=1$ for all $x\in X$.
Let $Y\subseteq X$ be a non-empty subset, and let $kY\subseteq kX$ be the subspace of $kX$ spanned by the elements of $Y$. Then you should be able to prove without much difficulty that $kY$ is a subcoalgebra of $kX$.
In the same spirit, it is easy to see that the $1$-dimensional subcoalgebras of a coalgebra $C$ are in bijection with the non-zero elements $g\in C$ such that $\Delta(g)=g\otimes g$.