Let $V \subset \mathcal{P}$ is given, where $\mathcal{P}$ is a set of all polynomials over $\mathbb{C}$. Does set of all vectors $x \in V$ such that $x(t) \geq 0$ for $t \in (0, 1)$ constitute a vector space?
Counterexample: consider some polynomial $s(t)$ which obeys the definiton of that set. If we take $\alpha = -1$, and make $\alpha\cdot s(t) = p(t)$ then polynomial $p(t) < 0$ for $t \in (0, 1)$ and thus set does not constitute a vector space, because it is not closed under multiplication.
Is my counterexample correct?
Your answer is essentially right, but it can be improved. First, you wrote that $p(t) < 0$ for $t \in (0,1)$; this is not true. Rather, $p(t) \le 0$ for $t \in (0,1)$. In particular, consider $s(t) = 0$ for all $t$: then your argument no longer works!
To fix this problem: consider some nonzero polynomial $s(t)$ in $V$, i.e., a polynomial such that $s(t_0) > 0$ for some $t_0$. Then $p(t_0) = -s(t_0) < 0$, so $p \not \in V$.
Second—and this is quite minor—it would be nice to say that $V$ is not closed under scalar multiplication.