Let $\mathfrak{g}$ be a Lie algebra and $M(\lambda)$ its Verma module to some weight $\lambda$. Taking the quotient of the Verma module with its unique maximal graded submodule $I(\lambda)$ we find a simple $\mathfrak{g}$-module $L(\lambda) = M(\lambda)/I(\lambda)$. If we consider the BGG category $\mathcal{O}$, this procedure, in fact, gives us all of the simple objects in $\mathcal{O}$.
Now, I'm wondering about the following: Let $V$ be any module in $\mathcal{O}$ (it is then, in fact, automatically a highest weight module). Can $V$ be written as the quotient $M(\mu)/N$ for some weight $\mu$ and some ideal $N$? My intuition is telling me this should be possible, but I'm currently not seeing why.
Follow-up question: Is $\mathcal{O}$ semisimple?
A module $V$ is highest weight of highest weight $\lambda$ if there exists a weight vector $v_\lambda \in V_\lambda$ such that $e_i(v) = 0$ for all $i$, and $v_\lambda$ generates $V$ as a $\mathfrak{g}$-module. (This is just a definition of what a highest weight module is).
If $V$ is a highest weight module of highest weight $\lambda$, then there exists a surjection $M(\lambda) \twoheadrightarrow V$ taking the highest-weight vector $m_\lambda$ to $v_\lambda$. So Verma modules always surject onto highest-weight modules. (In fact, you can use this as a definition of what a Verma module is, via a universal property).
Not every module of category $\mathcal{O}$ is highest weight. For example, if $V$ and $W$ are highest-weight modules, then their direct sum $V \oplus W$ is not, since it is not generated by a single weight vector. For a more involved example, take an indecomposable highest-weight module with nontrivial composition factors, and take the BGG dual.
Category $\mathcal{O}$ is not semisimple, since the Verma modules $M(\lambda)$ for dominant integral weights $\lambda$ have proper submodules which are not summands. You can see this easily even in the case $\mathfrak{g} = \mathfrak{sl}_2$, by writing out the basis for such a Verma module.