On p. 148 of Fulton and Harris' book "Representation Theory: A First Course", they write that "Moreover, by the same token, the $V_\alpha$ that appear must form an unbroken string of numbers of the form $\beta$, $\beta+2$, $\ldots$, $\beta+2k$."
I believed this when I read it two weeks ago, but now I'm not so sure about it. They define the subspace $W = \bigoplus_{k \in \mathbb{Z}} V_{\alpha_0+2k}$, where $\alpha_0$ is any complex number appearing in the decomposition, and note that $W$ is invariant so it must be equal to $V$. That is fine, but I don't understand why some of these spaces $V_{\alpha_0+2 \ell}$ can't be zero.
Assuming that $V$ is an irreducible:
Let $\beta = \alpha_{0} + 2k$ for some positive $k$, and assume that $V_{\beta} = 0$ but $V_{\beta+2} \neq 0$, and let $v \in V_{\beta+2}$. Let $x = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$ and $y = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}$, and note that $[x,y] = h = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$.
Then $y \cdot v \in V_{\beta}$, so it must be $0$. That implies that the submodule of $V$ generated by $v$ lives in $V_{\beta+2}$ and higher weight spaces, so it can't be all of $V$, contradicting the fact that $V$ is irreducible.
That shows that once some $V_{\alpha_{0}+2k} = 0$ for some positive $k$, every higher weight space must be 0 as well.