As the title says.
I seem to be proving the opposite which I doubt is correct.
Proof:
Choose arbitrary $y\in N(A_1)$.
Then, by $N(A_1)=R(A_2^T)$, $y\in R(A_2^T)$ as well.
Thus, $$ Ay = (A_1y,\ A_2y)^T $$ $$ = (0,\ A_2y)^T $$ However, $y\in R(A_2^T)$ as well. If $y\neq 0$, then $N(A_1) \neq 0$ in which case $R(A_2^T) \not\ni \{0\}$ which is absurd.
Thus, it must be that $y=0$ which in turn implies $N(A_1)=0$.
However, this implies that $R(A_2^T)=0$, in which case $A_2=0$.
If $A_2=0$, $A$ must be singular.
We have $$ A = \pmatrix{A_1\\A_2} $$ Note that for a vector $x$, we have $$ Ax = \pmatrix{A_1x\\A_2x} $$ We note that $A_1 x = 0$ iff $x \in N(A_1)$. Similarly, $A_2 x = 0$ iff $x \in N(A_2) = R(A_2^T)^\perp = N(A_1)^\perp$. It follows that $$ N(A) = N(A_1) \cap N(A_2) = N(A_1) \cap N(A_1)^\perp = \{0\} $$ Since $A$ is a square matrix with a trivial kernel, we conclude that it is non-singular.