Regioselectivity of deprotonation of 1,4-dicarbonyl
Here, both the α-$\ce{CH2}$ and α-$\ce{CH3}$ groups have very similar acidities since they are both α to only one carbonyl group. [This is different compared to, for example, a 1,3-dicarbonyl such as acetylacetone, where one $\ce{CH2}$ group is α to two carbonyl groups and is therefore significantly more acidic.] In our substrate, the two protons differ very slightly in acidities, and in general, whether one deprotonates at the more or less substituted carbon can be controlled by reaction conditions.

The mechanism you are drawing is the first step of what is known as an aldol condensation. Because all the steps in this reaction are easily reversible, simple aldol condensations with $\ce{OH-}$ are typically under thermodynamic control, meaning that the most stable product is preferentially formed.

Therefore, let's consider what would happen if we deprotonated the $\ce{-CH2}$ group. The deprotonation itself is definitely a possibility, but what happens after that is quite implausible:



Once you form the enolate, the next step in an aldol condensation is the enolate attacking the other carbonyl group. In this case, it would lead to a three-membered ring, where there is a lot of angle strain and the substituents eclipse each other. The dehydration to form the α,β-unsaturated ketone is even less likely, because the cyclopropene ring thus formed would possess incredible angle strain.

On the other hand, deprotonation of the terminal $\ce{-CH3}$ group eventually leads to the formation of a five-membered ring after nucleophilic attack on the other carbonyl group. This is much more stable, and you can eliminate water via an E1cb-type mechanism to get a cyclopentenone.



So, you could certainly deprotonate the $\ce{-CH2}$ position. It is just that, after it gets deprotonated, nothing very good can come out of it. This deprotonation is said to be unproductive.