$\newcommand\sp{\hspace{1cm}}$
1
$\|Xv\|^2=\|X_mv_m+X_{m^c}v_{m^c}\|^2$
$=\|X_mv_m\|^2+\|X_{m^c}v_{m^c}\|^2+2\langle X_mv_m,X_{m^c}v_{m^c}\rangle$
$=\|X_mv_m\|^2+\|X_{m^c}v_{m^c}\|^2+2v_{m^c}^TX_{m^c}^TX_mv_m$
$\ge \|X_mv_m\|^2-2|v_{m^c}^TX_{m^c}^TX_mv_m|$
2
First, $\|X_mv_m\|^2=\sum\limits_{\substack{i=1}}^{n}(\sum\limits_{\substack{j\in m}}{(X_{i,j}v_j)^2}+\sum\limits_{k\in m\ne j\in m}{X_{i,j}v_jX_{i,k}v_k})$
$=\sum\limits_{j\in m}(v_j^2\sum\limits_{\substack{i=1}}^{n}{X_{i,j}^2})+\sum\limits_{j\in m\ne k\in m}(v_jv_k\sum\limits_{i=1}^{n}{X_{i,j}X_{i,k}})$
$=\sum\limits_{j\in m}v_j^2+\sum\limits_{j\in m\ne k\in m}(v_jv_k\langle X_{j},X_{k}\rangle)$
$\geq\sum\limits_{j\in m}v_j^2-\sum\limits_{j\in m\ne k\in m}|v_j| |v_k|\theta$
$\geq \|v_m\|^2-\theta |v_m|_1^2$
3
Consider $v\in \mathscr{C}(\beta)$.
$|v_m^TX_m^TX_{m^c}v_{m^c}|=|\sum\limits_{\substack{i=1}}^{n}[(\sum\limits_{\substack{j \in m}}{v_{j}X_{i,j}})(\sum\limits_{\substack{k \in m^c}}{v_{k}X_{i,k}})] |$
$=|\sum\limits_{\substack{i=1}}^{n}\sum\limits_{\substack{j \in m}}\sum\limits_{\substack{k \in m^c}}{v_{j}X_{i,j}X_{i,k}v_{k}}|$
$=|\sum\limits_{\substack{j \in m}}\sum\limits_{\substack{k \in m^c}}{v_{j}v_{ k}}\sum\limits_{\substack{i=1}}^{n}{X_{i,j}X_{i,k}}|$
$=|\sum\limits_{\substack{j \in m}}\sum\limits_{\substack{k \in m^c}}{v_{j}v_{ k}\langle X_j,X_k\rangle}| \sp \le \sp \theta\sum\limits_{\substack{j \in m}}\sum\limits_{\substack{k \in m^c}}{|v_{j}||v_{k}|}$
$\le \theta |v_{m^c}|_1|v_{m}|_1 \le 5\theta |v_m|_1^2$
4
$\kappa(\beta)= \underset{v\in\mathscr{C}(\beta)}{min}\bigg\{\frac{\sqrt{|m|} \|Xv\|}{|v_m|_1}\bigg\}$
Then, $\frac{|m|\|Xv\|^2}{|v_m|_1^2} \sp \ge \sp |m|\bigg( \frac{\|X_mv_m\|^2-10\theta|v_m|_1^2}{|v_m|_1^2}\bigg)$
$\ge \sp |m|\bigg( \frac{\|v_m\|^2-\theta|v_m|_1^2}{|v_m|_1^2}\bigg)-|m|10\theta$
$\ge \sp \frac{|m|\|v_m\|^2}{|v_m|_1^2}-|m|11\theta$
$\geq 1-11|m|\theta$
since $|v_m|_1\leq \sqrt{|m|}\|v_m\|$ by Cauchy Schwartz.