Consider two real matrices $A \in \mathbb{R}^{n \times n}$ and $B \in \mathbb{R}^{n \times m}$, the target is to establish an inequality between $$ \sigma_{\mathrm{max}}(\begin{bmatrix} A& B\end{bmatrix}) $$ and $$ \sigma_{\mathrm{max}}(A), \text{ }\sigma_{\mathrm{max}}(B). $$
My hypothesis is that $$ \sigma_{\mathrm{max}}(\begin{bmatrix} A& B\end{bmatrix}) \geqslant \max\{\sigma_{\mathrm{max}}(A), \sigma_{\mathrm{max}}(B)\}, $$ which is trivial when both $A$ and $B$ are scalars. However, it is not clear for me how to prove it in general matrix cases.
It turns out that the proof is rather simple: \begin{align} &\sigma_{\mathrm{max}}(\begin{bmatrix} A& B\end{bmatrix}) = \sqrt{\lambda_{\mathrm{max}}(AA^\top +BB^\top)} \\ &\geqslant \max\{\sqrt{\lambda_{\mathrm{max}}(AA^\top )}, \sqrt{\lambda_{\mathrm{max}}(BB^\top )}\}= \max\{\sigma_{\mathrm{max}}(A), \sigma_{\mathrm{max}}(B)\}. \end{align}
The proof you present is valid. Here's another.
$$ \begin{align} \sigma_{\max} \pmatrix{A & B} &= \max_{x \in \Bbb R^{m+n}, \|x\|=1} \left\|\pmatrix{A & B}x \right\| \\ &= \max_{\|x_1\|^2 + \|x_2\|^2 = 1} \left\|\pmatrix{A & B}\pmatrix{x_1\\x_2} \right\| \\ &\geq \max_{\|x_1\|^2 + \|x_2\|^2 = 1, \ x_2 = 0} \left\|\pmatrix{A & B}\pmatrix{x_1\\x_2} \right\| = \max_{\|x_1\| = 1} \|Ax\| = \sigma_{\max}(A). \end{align} $$ By a similar argument, $\sigma_{\max} \pmatrix{A & B} \geq \sigma_{\max}(B)$.