While mimicking the union bound in quantum systems, we land on the following conjecture but don't know how to prove this. Given any complex-valued $n\times m$ matrix $A$. A sub-matrix of $A$ is defined by two index subsets $I \subseteq [n], J \subseteq [m]$, $$ (A\vert_{I, J})_{i j} := \begin{cases} A_{ij} \quad\text{ if }i\in I, j\in J \\ 0 \quad \text{ otherwise. } \end{cases} $$ Now there are $k$ index sets pairs $I_t\subset [n], J_t\subset [m]$ for $1\le t\le k$. Suppose that for any $i\in [n], j\in [m]$, there always exists some $t$ such that $i\in I_t, j\in J_t$. (Notice that $t$ might not be unique.)

Does the following inequality always hold? $$ \|A\|_{\mathrm{op}} \le \sum_{1\le t\le k}\big\|A\vert_{I_t, J_t}\big\|_{\mathrm{op}} $$ where $\|\cdot\|_{\mathrm{op}}$ is the operator 2-norm, or equivalently, the maximum singular value.

(We posted the question in Math Stackexchange, but then realized that this should be a research level question. Thus we have deleted the old post and re-post here.)