[proofplan]
We prove the inclusion by testing an arbitrary vector $x\cdot v$ against every element of $\mathfrak h$. The module identity rewrites the action of $H\in\mathfrak h$ on $x\cdot v$ in terms of $x\cdot(H\cdot v)$ and $[H,x]\cdot v$. The definitions of $V_\lambda$ and $\mathfrak g_\alpha$ then show that $H$ acts on $x\cdot v$ by the scalar $(\lambda+\alpha)(H)$, which is exactly the defining condition for membership in $V_{\lambda+\alpha}$.
[/proofplan]
[step:Test the image of an arbitrary weight vector against $\mathfrak h$]
Let $v\in V_\lambda$ be arbitrary. We must show that $x\cdot v\in V_{\lambda+\alpha}$.
By definition of the weight space $V_{\lambda+\alpha}$, this means proving that for every $H\in\mathfrak h$,
\begin{align*}
H\cdot(x\cdot v)=(\lambda+\alpha)(H)(x\cdot v).
\end{align*}
Since $v\in V_\lambda$, for every $H\in\mathfrak h$ we have
\begin{align*}
H\cdot v=\lambda(H)v.
\end{align*}
Since $x\in\mathfrak g_\alpha$, for every $H\in\mathfrak h$ we have
\begin{align*}
[H,x]=\alpha(H)x.
\end{align*}
[guided]
Choose an arbitrary vector $v\in V_\lambda$. To prove the inclusion $x\cdot V_\lambda\subseteq V_{\lambda+\alpha}$, it is enough to prove that this one vector $x\cdot v$ lies in $V_{\lambda+\alpha}$.
The definition of $V_{\lambda+\alpha}$ is a simultaneous eigenvector condition for every element of $\mathfrak h$. Thus we must prove that for each $H\in\mathfrak h$,
\begin{align*}
H\cdot(x\cdot v)=(\lambda+\alpha)(H)(x\cdot v).
\end{align*}
Fix $H\in\mathfrak h$. Because $v\in V_\lambda$, the vector $v$ has weight $\lambda$, so
\begin{align*}
H\cdot v=\lambda(H)v.
\end{align*}
Because $x\in\mathfrak g_\alpha$, the element $x$ lies in the $\alpha$-root space relative to the same subalgebra $\mathfrak h$, so
\begin{align*}
[H,x]=\alpha(H)x.
\end{align*}
The key point is that a module action converts the Lie bracket in $\mathfrak g$ into the commutator of endomorphisms of $V$. Therefore
\begin{align*}
[H,x]\cdot v=H\cdot(x\cdot v)-x\cdot(H\cdot v).
\end{align*}
Rearranging this identity gives
\begin{align*}
H\cdot(x\cdot v)=x\cdot(H\cdot v)+[H,x]\cdot v.
\end{align*}
Now substitute the two scalar-action identities. Since the action is linear in the vector argument and in the [Lie algebra](/page/Lie%20Algebra) argument,
\begin{align*}
H\cdot(x\cdot v)=x\cdot(\lambda(H)v)+(\alpha(H)x)\cdot v.
\end{align*}
Thus
\begin{align*}
H\cdot(x\cdot v)=\lambda(H)(x\cdot v)+\alpha(H)(x\cdot v).
\end{align*}
Combining the two scalar terms gives
\begin{align*}
H\cdot(x\cdot v)=(\lambda+\alpha)(H)(x\cdot v).
\end{align*}
Since $H\in\mathfrak h$ was arbitrary, this verifies the defining condition for $x\cdot v\in V_{\lambda+\alpha}$. Since $v\in V_\lambda$ was arbitrary, every vector in $x\cdot V_\lambda$ belongs to $V_{\lambda+\alpha}$, and hence
\begin{align*}
x\cdot V_\lambda\subseteq V_{\lambda+\alpha}.
\end{align*}
[/guided]
[/step]
[step:Use the module identity to compute the shifted weight]
Fix $H\in\mathfrak h$. Since $V$ is a $\mathfrak g$-module, its action satisfies
\begin{align*}
[H,x]\cdot v=H\cdot(x\cdot v)-x\cdot(H\cdot v).
\end{align*}
Rearranging gives
\begin{align*}
H\cdot(x\cdot v)=x\cdot(H\cdot v)+[H,x]\cdot v.
\end{align*}
Substituting $H\cdot v=\lambda(H)v$ and $[H,x]=\alpha(H)x$, and using linearity of the module action, gives
\begin{align*}
H\cdot(x\cdot v)=x\cdot(\lambda(H)v)+(\alpha(H)x)\cdot v.
\end{align*}
Therefore
\begin{align*}
H\cdot(x\cdot v)=\lambda(H)(x\cdot v)+\alpha(H)(x\cdot v).
\end{align*}
Hence
\begin{align*}
H\cdot(x\cdot v)=(\lambda+\alpha)(H)(x\cdot v).
\end{align*}
Since this holds for every $H\in\mathfrak h$, the vector $x\cdot v$ belongs to $V_{\lambda+\alpha}$. Because $v\in V_\lambda$ was arbitrary, we conclude that
\begin{align*}
x\cdot V_\lambda\subseteq V_{\lambda+\alpha}.
\end{align*}
[/step]
