Assume we work with the operator 2-norm. We have $$\|A\|_2^2 = \sup_{\|x\|=1} \|Ax\|_2^2=\sup_{\|x\|=1} \sum_{j=1}^{m}\sum_{i=1}^{n}(a_{ji}x_i)^2.$$ Let $I_t=\{i^t_{1}, \dots i^t_{p(t)}\}$ and $J_t=\{j^t_{1}, \dots j^t_{r(t)}\}$. Hence \begin{equation} \sum_{1 \leq t \leq k} \|A_{I_t,J_t}\|_2^2=\sum_{1 \leq t \leq k}\ \sup_{\|x\|=1}\|A_{I_t,J_t}x\|^2_2= \sup_{\|x\|=1}\ \sum_{1 \leq t \leq k} \sum_{k=1}^{r(t)} \sum_{l=1}^{p(t)}(a_{j^t_k i^t_l}x_i)^2. \end{equation} Now since for any $i \in [n],\ j \in [m]$ there is a $t$ such that $i \in I_t,\ j \in J_t$, the triple sum must contain all terms $(a_{ji}x_i)$ for $i \in [n],\ j \in [m]$ at least once. There could be some extra terms, which are non negative, so we have $$\|A\|_2^2 \leq \sum_{1 \leq t \leq k} \|A_{I_t,J_t}\|_2^2 \leq \bigg({\sum_{1 \leq t \leq k} \|A_{I_t,J_t}\|_2 }\bigg)^2.$$ Now $\sqrt{\cdot}$ is monotone, thus by taking the square root we get the desired inequality.

Assume we work with the operator 2-norm. We have $$\|A\|_2^2 = \sup_{\|x\|=1} \|Ax\|_2^2=\sup_{\|x\|=1} \sum_{j=1}^{m}(\sum_{i=1}^{n}a_{ji}x_i)^2.$$ Let $I_t=\{i^t_{1}, \dots i^t_{p(t)}\}$ and $J_t=\{j^t_{1}, \dots j^t_{r(t)}\}$. Hence \begin{equation} \sum_{1 \leq t \leq k} \|A_{I_t,J_t}\|_2^2=\sum_{1 \leq t \leq k}\ \sup_{\|x\|=1}\|A_{I_t,J_t}x\|^2_2= \sup_{\|x\|=1}\ \sum_{1 \leq t \leq k} \sum_{k=1}^{r(t)} (\sum_{l=1}^{p(t)}a_{j^t_k i^t_l}x_i)^2. \end{equation} Now since for any $i \in [n],\ j \in [m]$ there is a $t$ such that $i \in I_t,\ j \in J_t$, the triple sum must contain all terms $(a_{ji}x_i)$ for $i \in [n],\ j \in [m]$ at least once. There could be some extra terms, which are non negative, so we have $$\|A\|_2^2 \leq \sum_{1 \leq t \leq k} \|A_{I_t,J_t}\|_2^2 \leq \bigg({\sum_{1 \leq t \leq k} \|A_{I_t,J_t}\|_2 }\bigg)^2.$$ Now $\sqrt{\cdot}$ is monotone, thus by taking the square root we get the desired inequality.