The question is in the title. I think yes, and my reasoning is as follows, but something feels fishy and I can't quite put my finger on it.
Let $f:C\to D$ be a coalgebra morphism. Suppose that $Y\subset D$ is a subcoalgebra, and define $X\equiv f^{-1}(Y)$. Then for any $x\in X$ we have $f(x) = y\in Y$ and $\Delta(x)\in X\otimes X$, since $$ (f\otimes f \circ \Delta)(x) = (\Delta\circ f)(x) = \Delta(y) \in Y\otimes Y $$
I quitely assumed that $(f\otimes f)^{-1}(Y\otimes Y) = f^{-1}Y\otimes f^{-1}Y$, but I think this might be wrong. Can someone provide a counterexample if it exists? (this is actually really really circular)
This question came to me when I tried to find an answer for this question by user Bipolar Minds.
EDIT: While I'm interested in the general truth value, an answer for finite-dimensional coalgebras over a field of characteristic 0 would be a good starting point.
No, it is not generally true that $X$ is a subcoalgebra.
For a simple counterexample, let $K$ be any field, and let $C = K\left[C_2\right]$ be the group algebra of the two-element cyclic group $C_2 = \left\{1,g\right\}$ (regarded as a Hopf algebra, thus a coalgebra). The counit $\varepsilon : C \to K$ of $C$ is a coalgebra homomorphism. Its kernel $\ker\varepsilon$ is spanned by $g - 1$, and is not a subcoalgebra (since $\Delta\left(g-1\right) = g\otimes g - 1\otimes 1$ is not a multiple of $\left(g-1\right) \otimes \left(g-1\right)$). But this kernel has the form $\varepsilon^{-1} \left(0\right)$ for the subcoalgebra $0$ of $K$. Thus, a preimage of a subcoalgebra is not always a subcoalgebra.