For coassociative dg-coalgebras over any field $k$ the answer is positive, because:
Let $C$ be a $\mathbb Z$-graded coalgebra and $D\subset C$ a finite-dimensional ungraded subcoalgebra (of the underlying ungraded coalgebra) of $C$. Let $D^{gr}\subset C$ denote the graded vector subspace spanned by all the grading components of the elements of $D$. Then $D\subset D^{gr}$ and $D^{gr}$ is a finite-dimensional graded subcoalgebra of $C$.
Let $(C,d)$ be a dg-coalgebra and $D\subset C$ be a finite-dimensional graded subcoalgebra of $C$. Set $D^{dg}=D+d(D)\subset C$. Then $D\subset D^{dg}$ and $D^{dg}$ is a finite-dimensional dg-subcoalgebra of $C$.
Using the observations 1. and 2. and the fact that any ungraded coassociative coalgebra is the union of its finite-dimensional subcoalgebras, one deduces the assertion that any $\mathbb Z$-graded dg-coalgebra is the union of its finite-dimensional dg-subcoalgebras.
Possible generalizations: One can replace a field $k$ by a Noetherian commutative ring $k$ and speak about subcoalgebras that are finitely generated as $k$-modules (instead of "finite-dimensional"). All the assertions remain true.
One cannot drop the coassociativity condition. Indeed, even for ungraded coalgebras over a field of characteristic $0$, there is an example of a infinite-dimensional Lie coalgebra $L$ having no nonzero finite-dimensional subcoalgebras. The Lie coalgebra $L$ is simplest described in terms of its dual topological Lie algebra structure (on a pro-finite-dimensional topological vector space): $L^*=\mathfrak g=k[[z]]\,d/dz$, the Lie algebra of vector fields on the formal disk.