Let $n=2$. Let $\{V_t\}$ be the set of all lines through the origin.
If you really want to index these lines with the positive reals, find a one-to-one correspondence between the set of all reals, plus the symbol $\infty$, and the set of all positive reals.
Clearly the union of the $V_t$ is $\mathbb{R}^2$, and none of the $V_t$ is $\mathbb{R^2}$.
Exactly the same example works for $\mathbb{R}^n$ for any $n\ge 2$.
The situation is very different if the $V_t$ are nested, that is, if $s implies that $V_s \subseteq V_t$. For then by some finite $t$, we will have $V_t=\mathbb{R}^n$. The argument is simple, and has nothing much to do with uncountability. There must be some integer $n_1$ such that $V_{n_1}$ has dimension $\ge 1$. But then there must be an $n_2>n_1$ such that $V_{n_2}$ has dimension $\ge 2$. And so on. Sooner or later, we must reach an integer $n_k$ such that $V_{n_k}=\mathbb{R}^n$.