(We have 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.)

While mimicing 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?
$$
\|A\|_{op} \le \sum_{1\le t\le k}\|A\vert_{I_t, J_t}\|_{op}
$$
where $\|\cdot\|_{op}$ is the operator 2-norm, or equivalently, the maximum singular value.