Let $W(t)$ a Wiener process. By the scaling propety of the Brownian motion we have that $$ \frac{1}{\sqrt{c}} W(ct) $$ is a Wiener process. Let $(\mathcal{F}_t)_{t \ge 0}$ the Filtration of the Wiener process and suppose $cs <t$. Then $W(cs)$ is $\mathcal{F}_t$ measurable
As $\frac{1}{\sqrt{c}} W(cs) \sim W(s)$ ($\sim$ here is equality in law) it looks to me that
$$ \operatorname{Cov}\left( c^{-\frac{1}{2}} W(cs), W(t) \right)= \operatorname{Cov}(W(s),W(t)) $$ But looking at this question (in particular one of the comments and the accepted answer) it looks like this is not the case.
Is this formula false? In that case what is $\operatorname{Cov}( c^{-\frac{1}{2}} W(cs), W(t) )$?
$\frac{1}{\sqrt c} W(cs)$ is a Brownian motion, but it isn't the same Brownian motion that you started with, so $\text{Cov}(\frac{1}{\sqrt c} W(cs),W(t)) \ne \text{Cov}(W(s),W(t))$. Instead, using standard rules of covariance and the fact that $cs < t$, we have \begin{align*} \text{Cov}(\frac{1}{\sqrt c} W(cs),W(t)) &= \frac{1}{\sqrt c} \text{Cov}(W(cs),W(t)) \\ &= \frac{1}{\sqrt c} cs \\ &= \sqrt{c} s \end{align*}