At a particular temperature, the $K_{\text{eq}}$ for the following reaction (yes, it's the auto-protolysis of water):
$$\ce{H2O(l) <=> H+(aq) + OH-(aq)}$$
will be constant. Note that $K_\mathrm{w}=K_{\text{eq}}\times[\ce{H2O}]=\ce{[H+(aq)][OH-(aq)]}$ is called the auto-protolysis constant of water. It is considered a constant because concentration of water $([\ce{H2O}])$ is assumed to be constant.
So, you can directly observe from here, that since $\ce{[H+(aq)][OH-(aq)]}=K_\mathrm{w}=\text{constant}$ at a particular temperature, if $\ce{[H+(aq)]}$ increases (i.e. $\mathrm{pH}$ decreases), $[\ce{OH-(aq)}]$ must decrease (i.e. $\mathrm{pOH}$ must increase).
Your interpretation of the Le Chatelier's principle is also correct. In fact, it is actually what is happening behind the scenes. When you add more acid (assuming a 100% ionized acid), the $Q$ value of the auto-protolysis of water increases, causing the reaction to shift backward. However, the backward shift is unable to decrease the concentration of $\ce{H+}$ ions as much as it decreases the concentration of $\ce{OH-}$ ions. See explanation below:
Consider adding $\pu{0.01M}$ $\ce{HCl}$ to pure water (at $\pu{25^\circ C}$) at $t=0$. This is what happens till achieving equilibrium at $t=t_0$:
$$
\begin{array}{ccc}
&\ce{H2O(l) &<=>& H+(aq)&+&OH-(aq)}\\\hline
-&-&&\pu{10^-7M}&&\pu{10^-7M}\\
t=0&-&&\pu{(10^-7+10^-2)M}&&\pu{10^-7M}\\
t=t_0&-&&(10^{-7}+10^{-2}-x)~\pu{M}&&(10^{-7}-x)~\pu{M}\\
\end{array}
$$
Now, if you actually solve for $x$ at $t=t_0$, you'll find $x$ to be of the order $10^{-8}$ or below. Notice that, for the $\ce{H+}$ ions, this means a decreases in concentration from $\approx\pu{0.01M}$ at $t=0$ to $\pu{0.0099999M}$ at $t=t_0$. This a hardly noticeable change. However, it is a catastrophe for the $\ce{OH-}$ ions, whose concentration would probably be reduced, by a factor of more than ten thousand, from the initial $\pu{10^-7 M}$! The change in $\mathrm{pOH}$ is, hence, much more pronounced.
