= Solution
Set $C=\sup_k\|\mu_k\|$. Given $f\in C(\mathbb T)$, choose $g\in X$ with $\|f-g\|_\infty<\varepsilon$. Then
$$
\left|\int f\,d\mu_k-\int f\,d\mu\right|\leq(C+\|\mu\|)\varepsilon+\left|\int g\,d\mu_k-\int g\,d\mu\right|.
$$
Take the upper limit in $k$ and then $\varepsilon\downarrow0$. Thus \b[convergence on the dense test family suffices]; no assumption that $X$ is a vector space is needed.
For nonempty compact $K$, put $f_j(x)=\max\{0,1-j\operatorname{dist}(x,K)\}$. These <continuous functions> decrease to $\mathbf1_K$ and satisfy $\mu_k(K)\leq\int f_j\,d\mu_k$. Consequently $\limsup_k\mu_k(K)\leq\int f_j\,d\mu$ for every $j$. <Dominated convergence> for the finite <positive measure> $\mu$ yields
$$
\boxed{\limsup_k\mu_k(K)\leq\mu(K).}
$$
The empty-set case is immediate. Positivity is essential for the comparison with these cutoff functions.
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