Let's see how far we can go with this, using a bit naive/informal reasoning:
Define $(\hat \theta_n - \theta)^2 \equiv \Delta_n$. Then
$${\rm consistency}: \Pr(|\hat \theta_n - \theta| < \epsilon') \to 1 \implies \Pr(|\hat \theta_n - \theta|^2 > \epsilon'^2) \to 0,$$
$$\implies \Pr(\Delta_n \geq \epsilon) \to 0. \tag{1}$$
Also,
$$MSE \equiv E(\hat \theta_n - \theta)^2 = E(\Delta_n). \tag{2}$$
For some $\epsilon >0$, decompose the expected value
$$E(\Delta_n) = E(\Delta_n \mid \Delta_n < \epsilon)\cdot\Pr(\Delta_n < \epsilon) + E(\Delta_n \mid \Delta_n \geq \epsilon)\cdot\Pr(\Delta_n \geq \epsilon).$$
We also have by basic properties of the expected value,
$$E(\Delta_n \mid \Delta_n < \epsilon) \leq \epsilon,$$
and since $\Pr$ is bounded by unity
$$MSE = E(\Delta_n) \leq \epsilon + E(\Delta_n \mid \Delta_n \geq \epsilon)\cdot\Pr(\Delta_n \geq \epsilon) \tag{3}$$
As sample size $n$ increases, we will eventually be able to choose smaller and smaller $\epsilon$. Also, we know from the consistency property that $\Pr(\Delta_n \geq \epsilon)$ will tend to zero. So the weak consistency property certainly pushes for the MSE to reduce in value.
More over, if $E(\Delta_n \mid \Delta_n \geq \epsilon)$ is bounded, the second term will certainly go to zero.
But, in the thread the OP links to, we see an example where in an analogous situation (with a mixture estimator), the corresponding part of the MSE stabilizes to a non-zero value.
In such a case also the MSE falls, although it does not reach zero. This and other such examples have certainly an artificial flavor, and their main purpose is to show why MSE-consistency is a stronger property.
But for most, if not all practical cases, we will see the MSE fall.
