Let $V$ be an irreducible $\mathbb{C}S_n$-module, and $L_i$ the $i$th Jucys-Murphy element of $\mathbb{C}S_n$. A vector $\lambda=(\lambda_1,\dots,\lambda_n)$ is a weight of $V$ if there is a simulteanous eigenspace $V_\lambda$ in $V$ where $L_i$ acts as $\lambda_i$. Such $V_\lambda$ are one dimensional and spanned by some vector $v_\lambda$.
If $\mu=s_i\lambda=(\lambda_1,\dots,\lambda_{i-1},\lambda_{i+1},\lambda_i,\lambda_{i+2},\dots,\lambda_n)$, but $\lambda_{i+1}=\lambda_i\pm 1$, why is $\mu$ not a weight of $V$?
This is part of Corollary 2.2.3 of Kleshchev's Linear and Projective Representations of Symmetric Groups. He proves that $L(\lambda_i,\lambda_{i+1})$ is a one dimensional irreducible $\mathcal{H}_2$-module, and says it follows from the classification of $\mathcal{H}_2$-modules.
Here $\mathcal{H}_2=\langle x,y,s:xy=yx,\ s^2=1,\ sx=ys-1\rangle$, and if $\lambda_{i+1}=\lambda_i+1$, then $x,y,s$ act by $\lambda_i,\lambda_{i+1},-1$, respectively on a certain basis vector of $L(\lambda_i,\lambda_{i+1})$. But I don't see how this connects with $\mu$ and why it's not a weight.
Note that the module $M$ he defines, $$M := \operatorname{Hom}_{S_{i-1}}(V(\lambda_1, \ldots, \lambda_{i-1}), V(\lambda_1, \ldots, \lambda_{i+1})),$$ is the space of copies of $V(\lambda_1, \ldots, \lambda_{i-1})$ in $\operatorname{res}^{S_{i+1}}_{S_{i-1}} V(\lambda_1, \ldots, \lambda_{i+1})$. In particular, if $s_i \lambda$ is a weight, then $M$ will "see" it.
He claims $M$ is irreducible as a $\mathcal B_i := \langle L_i, s_i, L_{i+1}\rangle$-module (using a technical theorem from the introduction), and is isomorphic to $N := \mathcal B_i v_\lambda$ (this isomoprhism is induced by the homomorphism $M \to N$ given by $f \mapsto f(v_\lambda)$).
Once you believe these two claims, it falls out of the classification of irreducible representations of $\mathcal H_2 \twoheadrightarrow \mathcal B_i$.
Hope that helps!