Confusion with the narrow and weak* convergence of measures

The mistake is in

This subsequence trivially satisfies $\mu_n(X) = 1 \to 1 = \mu(X)$.

Consider a sequence $(x_n)$ of points in $X$ converging to $\infty$ (that is, for every compact subset $K$ of $X$ the set $\{n \in \mathbb{N} ; x_n \in K\}$ is finite), and let $\mu_n$ be the point mass in $x_n$ (or $\mu_n = \delta_{x_n}$).

Then you have $\mu_n \to 0$ in the weak$^{\ast}$ topology since $f(x_n) \to 0$ for every $f \in C_0(X)$, and $0$ clearly isn't a probability measure.

The tightness ensures that such things cannot happen. In the example, the mass "escapes to infinity", which tightness prohibits.