The answer, in general is negative. That is, there exist continuous maps $T:X \to X$ of the type you describe and Borel probability measures $\mu$ such that, for small $\epsilon$, the set $B_\epsilon(\mu)$ is empty. For instance, this happens whenever $T$ is uniquely ergodic [1] (The canonical example of that is an irrational rotation on the circle.)

Indeed, if $T$ is uniquely ergodic, then there is a unique invariant Borel probability measures $\nu_T$ on $X$. Observe that for each $x$ the measures in $V(x)$ are all $T$-invariant, so $V(x)=\{\nu_T\}$. If $\mu \ne \nu_T$ then for small enough $\epsilon$, we have $\nu_T \notin N_\epsilon(\mu)$, so $B_\epsilon(\mu)$ is empty.

[1] https://en.wikipedia.org/wiki/Ergodicity#Unique_ergodicity

The answer, in general, is negative. That is, there exist continuous maps $T:X \to X$ of the type you describe and Borel probability measures $\mu$ such that, for small $\epsilon$, the set $B_\epsilon(\mu)$ is empty. For instance, this happens whenever $T$ is uniquely ergodic [1] (The canonical example of that is an irrational rotation on the circle.)

Indeed, if $T$ is uniquely ergodic, then there is a unique $T$-invariant Borel probability measure $\nu_T$ on $X$. Observe that for each $x$ the measures in $V(x)$ are all $T$-invariant, so $V(x)=\{\nu_T\}$. If $\mu \ne \nu_T$ then for small enough $\epsilon$, we have $\nu_T \notin N_\epsilon(\mu)$, so $B_\epsilon(\mu)$ is empty.

[1] https://en.wikipedia.org/wiki/Ergodicity#Unique_ergodicity