If $\mathfrak{g}$ is a semisimple Lie algebra, with Cartan subalgebra $\mathfrak{h}$, and irreducible representation $(V, \rho)$, where $V$ has a non-zero weight-space, then $V$ is the direct sum of its weight spaces.
I see why is that the intersection of two distinct weight spaces is $\{0\}$, indeed let $0 \neq v \in V_\lambda \cap V_\mu, \ $ then: $\forall \ h \in \mathfrak{h}, \ \ \lambda(h)\cdot v = h \cdot v =\mu(h)\cdot v \Rightarrow \forall \ h \in \mathfrak{h},\ (\lambda-\mu)(h)=0, \ \text{ as } \ v \neq 0 \ \Rightarrow \ \mu =\lambda.$
But it's not clear to me why the sum of the weight spaces must equal the entire space.
You can express $\mathfrak g$ itself as $\mathfrak h\oplus\bigoplus_{\alpha\in R}\mathfrak g_\alpha$, whre $R$ is the set of roots.
Now, consider the sum $S$ of the weight spaces; you want to prove that $S=V$. You know that $S\ne\{0\}$ , since you are working over $\Bbb C$, which is algerbaically closed, and therefore the set of wights is not empty. On the other hand, by the definition of $S$, if $H\in\mathfrak h$ and $v\in S$, then $Hv\in S$. And if $X\in\mathfrak g_\alpha$, for some root $\alpha$, and if $v\in V_\beta$, for some weight $\beta\in\mathfrak h^*$, then $Xv\in V_{\alpha+\beta}\in S$. But then $S$ is a submodule which is not $\{0\}$. So, since $V$ is irreducible, $S=V$.