To do it by bare hands, I'd argue as follows. Given $x$ with the property that $\omega_{std}(x,y) = 0$ for all $y$, express $x$ in the basis $\partial q_i, \partial p_j$. So $x = a^i \partial q_i + b^j \partial p_j$ (with the Einstein summation convention) for some choices of $a^i$ and $b^j$. Our goal is to show that each $a^i = b^j = 0$.
Well, set $y = \partial p_k$. Then we get \begin{align*} 0 &= \omega_{std}(x,\partial p_k)\\ &= \sum (dq^i \wedge dp^i)(x,\partial p_k)\\ &= \sum dq^i(x)dp^i(\partial p_k) - dq^i(\partial p_k) dp^i(x) \end{align*}
Now, the basis $\{dq^i, dp^j\}$ is dual to $\{\partial q_i, \partial p_j\}$, so this sum isn't too hard to evaluate. The second term of the sum is identically $0$ since $dq^i(\partial p_k) = 0$. The first is $0$ unless $i = k$ since $dp^i(\partial p_k) = 0$ for $i\neq k$ and equals $1$ when $i = k$.
So, the sum reduced to a single term: $0 = dq^k(x)dp^k(\partial p_k) = dq^k(x)$
Finally, to evaluate $dq^k(x)$, recall $x = a^i \partial q_i + b^j \partial p_j$. So \begin{align*} dq^k(x) &= dq^k(a^i \partial q_i + b^j \partial p_j)\\ &= a^i dq^k(\partial q_i) + b^j dq^k(\partial p_j)\\ &=a^k\end{align*}
where the last step uses the fact that you're working in a dual basis. Putting this altogether, we just showed $0 = a^k$. Doing this for all $k$, and then using $y = \partial q_k$ to deal with the $b^j$ shows $x = 0$.