Let's write the $N$-particle space $H^{\otimes N}$ as $H^{\otimes N-1}\otimes H$ so that we have the basis $$ \lvert \bar{s},m_{\bar{s}}; m_N\rangle$$ where $\bar{s}$ is the total spin for the $N-1$ particles and $m_N$ the $z$-spin of the $N$-th particle (since there can be more than one copy of any given $\bar{s}$ representation, we would also have to have a marker for that duplicity. I'm omitting it here for convenience of notation as I don't think it changes anything about the argument). Expressing a state of definite total spin $s$ in terms of this is a standard application of Clebsch-Gordan coefficients: $$ \lvert s,m_s\rangle = \sum_{\bar{s}}\sum_{m_{\bar{s}}} \sum_{m_N}C^{sm_s}_{\bar{s}m_{\bar{s}}\frac{1}{2}m_N}\lvert \bar{s},m_{\bar{s}}; m_N\rangle,\tag{1}$$ where the coefficients are only non-zero for $\lvert \bar{s}-\frac{1}{2}\rvert\leq s\leq \bar{s}+\frac{1}{2}$ and $m_s = m_{\bar{s}} + m_N = m_{\bar{s}} \pm\frac{1}{2}$.

The question is now when $\langle s_1, m_{s_1}\lvert \sigma^x_N \lvert s_2,m_{s_2}\rangle$ is non-zero. In the expansion (1), the action of $\sigma^x_N$ is just to flip $m_N$ but doesn't touch the $\bar{s}$ part. Since the $ \lvert \bar{s},m_{\bar{s}}; m_N\rangle$ are an orthonormal basis, that means this overlap can only be non-zero when the two $\lvert s_i,m_{s_i}\rangle$ have at least one non-zero $\bar{s}$ in common. This can only happen when that $\bar{s}$ fulfills both $\lvert \bar{s}-\frac{1}{2}\rvert\leq s_1\leq \bar{s}+\frac{1}{2}$ and $\lvert \bar{s}-\frac{1}{2}\rvert\leq s_2\leq \bar{s}+\frac{1}{2}$. Write $s_2 = s_1 + x$, then $$ \lvert \bar{s}-\frac{1}{2}\rvert \leq s_1\leq \bar{s}+\frac{1}{2} - x,$$ which means $x$ is at most 1, since $\lvert \bar{s}-\frac{1}{2}\rvert$ and $\bar{s} + \frac{1}{2}$ differ at most by 1. This is what we wanted to show.

Let's write the $N$-particle space $H^{\otimes N}$ as $H^{\otimes N-1}\otimes H$ so that we have the basis $$ \lvert \bar{s},m_{\bar{s}}; m_N\rangle$$ where $\bar{s}$ is the total spin for the $N-1$ particles and $m_N$ the $z$-spin of the $N$-th particle (since there can be more than one copy of any given $\bar{s}$ representation, we would also have to have a marker for that duplicity. I'm omitting it here for convenience of notation as I don't think it changes anything about the argument [Edit: confirmed]). Expressing a state of definite total spin $s$ in terms of this is a standard application of Clebsch-Gordan coefficients: $$ \lvert s,m_s\rangle = \sum_{\bar{s}}\sum_{m_{\bar{s}}} \sum_{m_N}C^{sm_s}_{\bar{s}m_{\bar{s}}\frac{1}{2}m_N}\lvert \bar{s},m_{\bar{s}}; m_N\rangle,\tag{1}$$ where the coefficients are only non-zero for $\lvert \bar{s}-\frac{1}{2}\rvert\leq s\leq \bar{s}+\frac{1}{2}$ and $m_s = m_{\bar{s}} + m_N = m_{\bar{s}} \pm\frac{1}{2}$.

The question is now when $\langle s_1, m_{s_1}\lvert \sigma^x_N \lvert s_2,m_{s_2}\rangle$ is non-zero. In the expansion (1), the action of $\sigma^x_N$ is just to flip $m_N$ but doesn't touch the $\bar{s}$ part. Since the $ \lvert \bar{s},m_{\bar{s}}; m_N\rangle$ are an orthonormal basis, that means this overlap can only be non-zero when the two $\lvert s_i,m_{s_i}\rangle$ have at least one non-zero $\bar{s}$ in common. This can only happen when that $\bar{s}$ fulfills both $\lvert \bar{s}-\frac{1}{2}\rvert\leq s_1\leq \bar{s}+\frac{1}{2}$ and $\lvert \bar{s}-\frac{1}{2}\rvert\leq s_2\leq \bar{s}+\frac{1}{2}$. Write $s_2 = s_1 + x$, then $$ \lvert \bar{s}-\frac{1}{2}\rvert \leq s_1\leq \bar{s}+\frac{1}{2} - x,$$ which means $x$ is at most 1, since $\lvert \bar{s}-\frac{1}{2}\rvert$ and $\bar{s} + \frac{1}{2}$ differ at most by 1. This is what we wanted to show.

Let's write the $N$-particle space $H^{\otimes N}$ as $H^{\otimes N-1}\otimes H$ so that we have the basis $$ \lvert \bar{s},m_{\bar{s}}; m_N\rangle$$ where $\bar{s}$ is the total spin for the $N-1$ particles and $m_N$ the $z$-spin of the $N$-th particle. Expressing a state of definite total spin $s$ in terms of this is a standard application of Clebsch-Gordan coefficients: $$ \lvert s,m_s\rangle = \sum_{\bar{s}}\sum_{m_{\bar{s}}} \sum_{m_N}C^{sm_s}_{\bar{s}m_{\bar{s}}\frac{1}{2}m_N}\lvert \bar{s},m_{\bar{s}}; m_N\rangle,\tag{1}$$ where the coefficients are only non-zero for $\lvert \bar{s}-\frac{1}{2}\rvert\leq s\leq \bar{s}+\frac{1}{2}$ and $m_s = m_{\bar{s}} + m_N = m_{\bar{s}} \pm\frac{1}{2}$.

The question is now when $\langle s_1, m_{s_1}\lvert \sigma^x_N \lvert s_2,m_{s_2}\rangle$ is non-zero. In the expansion (1), the action of $\sigma^x_N$ is just to flip $m_N$ but doesn't touch the $\bar{s}$ part. Since the $ \lvert \bar{s},m_{\bar{s}}; m_N\rangle$ are an orthonormal basis, that means this overlap can only be non-zero when the two $\lvert s_i,m_{s_i}\rangle$ have at least one non-zero $\bar{s}$ in common. This can only happen when that $\bar{s}$ fulfills both $\lvert \bar{s}-\frac{1}{2}\rvert\leq s_1\leq \bar{s}+\frac{1}{2}$ and $\lvert \bar{s}-\frac{1}{2}\rvert\leq s_2\leq \bar{s}+\frac{1}{2}$. Write $s_2 = s_1 + x$, then $$ \lvert \bar{s}-\frac{1}{2}\rvert \leq s_1\leq \bar{s}+\frac{1}{2} - x,$$ which means $x$ is at most 1, since $\lvert \bar{s}-\frac{1}{2}\rvert$ and $\bar{s} + \frac{1}{2}$ differ at most by 1. This is what we wanted to show.

Let's write the $N$-particle space $H^{\otimes N}$ as $H^{\otimes N-1}\otimes H$ so that we have the basis $$ \lvert \bar{s},m_{\bar{s}}; m_N\rangle$$ where $\bar{s}$ is the total spin for the $N-1$ particles and $m_N$ the $z$-spin of the $N$-th particle (since there can be more than one copy of any given $\bar{s}$ representation, we would also have to have a marker for that duplicity. I'm omitting it here for convenience of notation as I don't think it changes anything about the argument). Expressing a state of definite total spin $s$ in terms of this is a standard application of Clebsch-Gordan coefficients: $$ \lvert s,m_s\rangle = \sum_{\bar{s}}\sum_{m_{\bar{s}}} \sum_{m_N}C^{sm_s}_{\bar{s}m_{\bar{s}}\frac{1}{2}m_N}\lvert \bar{s},m_{\bar{s}}; m_N\rangle,\tag{1}$$ where the coefficients are only non-zero for $\lvert \bar{s}-\frac{1}{2}\rvert\leq s\leq \bar{s}+\frac{1}{2}$ and $m_s = m_{\bar{s}} + m_N = m_{\bar{s}} \pm\frac{1}{2}$.

The question is now when $\langle s_1, m_{s_1}\lvert \sigma^x_N \lvert s_2,m_{s_2}\rangle$ is non-zero. In the expansion (1), the action of $\sigma^x_N$ is just to flip $m_N$ but doesn't touch the $\bar{s}$ part. Since the $ \lvert \bar{s},m_{\bar{s}}; m_N\rangle$ are an orthonormal basis, that means this overlap can only be non-zero when the two $\lvert s_i,m_{s_i}\rangle$ have at least one non-zero $\bar{s}$ in common. This can only happen when that $\bar{s}$ fulfills both $\lvert \bar{s}-\frac{1}{2}\rvert\leq s_1\leq \bar{s}+\frac{1}{2}$ and $\lvert \bar{s}-\frac{1}{2}\rvert\leq s_2\leq \bar{s}+\frac{1}{2}$. Write $s_2 = s_1 + x$, then $$ \lvert \bar{s}-\frac{1}{2}\rvert \leq s_1\leq \bar{s}+\frac{1}{2} - x,$$ which means $x$ is at most 1, since $\lvert \bar{s}-\frac{1}{2}\rvert$ and $\bar{s} + \frac{1}{2}$ differ at most by 1. This is what we wanted to show.