Of course, there are many ways of metrizing the weak topology on $\mathcal M(\Omega)$ by using various tools of functional analysis. However, as it has already been pointed out by Dan, the most natural way is to use the transportation metric on the space of measures. [It is much more natural than the Prokhorov metric. I don't want to go into historical details here - they can be easily found elsewhere, but I insist that the transportation metric should really be related with the names of Kantorovich (in the first place) and his collaborator Rubinshtein]. Dan gives its dual definition in terms of Lipschitz functions, however its "transport definition" is actually more appropriate here. Let me remind it.
Given two probability measures $\mu_1,\mu_2$ on $\Omega$
$$
\overline d(\mu_1,\mu_2) = inf_M \int d(x_1,x_2) dM(x_1,x_2) \;,
$$
where $d$ is the original metric on $\Omega$, and the infimum (which is in fact attained) is taken over all probability measures $M$ on $\Omega\times\Omega$ whose marginals ($\equiv$ coordinate projections) are $\mu_1$ and $\mu_2$. One should think about such measures as "transportation plans" between distributions $\mu_1$ and $\mu_2$, while the integral in the RHS of the definition is the "cost" of the plan $M$.
It is obvious that the above definition makes sense not just for probability measures, but for any two positive measures $\mu_1,\mu_2$ with the same mass. Moreover, $\overline d(\mu_1,\mu_2)$ actually depends on the difference $\mu_1-\mu_2$ only, so that one can think about it as a "weak norm"
$$
|||\mu_1-\mu_2||| = \overline d(\mu_1,\mu_2)
$$
of the signed measure $\mu_1-\mu_2$ (clearly, it is homogeneous with respect to multiplication by scalars).
Let now $\mu=\mu_1-\mu_2$ be an arbitrary signed measure, where $\mu_1,\mu_2$ are the components of its Hahn decomposition. The only reason why the definition of the weak norm does not work in this situation is that the measure $\mu$ need not to be "balanced" in the sense that the total masses $\|\mu_1\|$ and $\|\mu_2\|$ need not be the same any more. However, this can be easily repaired in the following way: extend the original space $\Omega$
to a new metric space $\Omega'$ by adding to it an "ideal point" $o$ and putting $d(\omega,o)=1$ for any $\omega\in\Omega$. Then the measure
$$
\mu'=\mu - (\|\mu_1\|-\|\mu_2\|)\delta_o \;,
$$
where $\delta_o$ is the unit mass at the point $o$, is now balanced, so that $|||\mu'|||$ is well defined. Therefore, one can extend the definition of the weak norm $|||\cdot|||$ to arbitrary signed measures $\mu$ by putting
$$
|||\mu|||=|||\mu'||| \;.
$$
It is now easy to see that the distance $|||\mu_1-\mu_2|||$, where $\mu_1,\mu_2$ are two arbitrary signed measures, metrizes the weak topology on $\mathcal M(\Omega)$.
First, I interpret your condition that "$X$ has no isolated points" in the following ways: First, every ball $B(x, \varepsilon)$ has non-empty interior. This means, in particular, that we can find arbitrarily fine partitions of $X$ with sets that each have non-empty interior. And second, if we have $X = \Omega_1 \stackrel{.}{\cup} \Omega_2$ for two non-empty sets $\Omega_1, \Omega_2$, then $\inf_{x_1 \in \Omega_1, x_2 \in \Omega_2} d(x_1, x_2) = 0$. I am not entirely sure if these conditions can be stated in simpler terms, but they don't seem unreasonable to me.
If this is what you had in mind (it's certainly satisfied for $[0, 1]$ or convex and compact subsets of $\mathbb{R}^d$), then I believe the two notions of convergence are indeed equivalent, see the proposed proof below. It's rather lengthy, so I hope I didn't miss anything. Please let me know if the proof seems incomplete.
Let $\mu \in \mathcal{P}(X)$ be fully supported and $(\mu_k)_{k \in \mathbb{N}}$ be a sequence in $\mathcal{P}(X)$ such that $\mu_k$ converges weakly to $\mu$. In particular, $d_P(\mu_k, \mu) \rightarrow 0$ for the Prokhorov metric $d_P$.
Let $\mathcal{F}_1, \mathcal{F}_2, ...$ be a sequence of refinements of partitions of $X$, i.e., $\mathcal{F}_N = \{\Omega_{N, 1}, \dots, \Omega_{N, N}\}$, $X = \stackrel{.}{\cup}_{i=1}^N \Omega_{N, i}$, such that $\delta(N) := \max_{i=1, \dots, N} \max_{x, y \in \Omega_{N, i}}d(x, y) \rightarrow 0$ for $N \rightarrow 0$, each $\Omega_{N, i}$ has non-empty interior and choose some fixed $x_{N, i} \in \Omega_{N, i}$. Note that by the "connectedness condition" on X, for any $J \subsetneq \{1, \dots, N\}$, it holds $\min_{j \in J} \min_{i \not\in J} d(x_i, x_j) \leq 2 \delta(N)$.
Define $\bar\mu^N := \sum_{i=1}^N \delta_{x_{N, i}} \mu(\Omega_{N, i})$. It is clear that $W_\infty(\mu, \bar\mu^N) \leq \delta(N)$. The fact that the partitions are refinements implies, in particular, that $\bar\mu^{N_2}(\Omega_{N_1, i}) = \bar\mu^{N_1}(\Omega_{N_1, i})$ for $N_2 \geq N_1$.
Define $w(N) := \min_{i=1, \dots, N} \mu(\Omega_{N, i})$, which is positive by assumption on $\mu$.
Note that for any $N_1, N_2 \in \mathbb{N}$ with $N_2 \geq N_1$ and any $A \subset X$ open such that $\bar\mu^{N_2}(A^{3 \delta(N_1)}) \neq 1$, it holds
$$
\bar\mu^{N_2}(A^{4 \delta(N_1)}) \geq \bar\mu^{N_2}(A^{2 \delta(N_1)}) + w(N_1),
$$
the reason being as follows: $\bar\mu^{N_2}(A^{3 \delta(N_1)}) \neq 1$ implies that there exists some $i \in \{1, \dots, N_1\}$ such that $\Omega_{N_1, i} \cap A^{2 \delta(N_1)} = \emptyset$ (since otherwise $A^{3 \delta(N_1)} = X$). In particular, by connectedness of $X$ there exists such an $i$ such that $\Omega_{N_1, i} \subset A^{4 \delta(N_1)}$. I.e.,
\begin{align}
\bar\mu^{N_2}(A^{4 \delta(N_1)}) &\geq \bar\mu^{N_2}(A^{2 \delta(N_1)}) + \bar\mu^{N_2}(\Omega_{N_1, i})\\ &= \bar\mu^{N_2}(A^{2 \delta(N_1)}) + \bar\mu^{N_1}(\Omega_{N_1, i})\\ &\geq \bar\mu^{N_2}(A^{2 \delta(N_1)}) + w(N_1).
\end{align}
Now for $\varepsilon > 0$, we choose $N_1$ large enough so that $\delta(N_1) < \varepsilon$. Let $r(N_1) := \min\{2 \delta(N_1), w(N_1)\}$. Choose $N_2, k$ large enough so that $d_P(\bar\mu^{N_2}, \mu_k) \leq r(N_1)$.
Let $A \subset X$ open. Either $\bar\mu^{N_2}(A^{3 \delta(N_1)}) = 1$, and hence $\bar\mu^{N_2}(A^{4 \delta(N_1)}) \geq \mu_k(A)$ holds trivially. Or, $\bar\mu^{N_2}(A^{3 \delta(N_1)}) < 1$ and hence by the above
$$
\bar\mu^{N_2}(A^{4 \delta(N_1)}) \geq \bar\mu^{N_2}(A^{2 \delta(N_1)}) + w(N_1) \geq \bar\mu^{N_2}(A^{r(N_1)}) + r(N_1) \geq \mu_k(A),
$$
where the last step follows since $d_P(\bar\mu^{N_2}, \mu_k) \leq r(N_1)$. Therefore, $W_\infty(\bar\mu^{N_2}, \mu_k) \leq 4 \delta(N_1) \leq 4 \varepsilon$. By the triangle inequality, $W_\infty(\mu, \mu_k) \leq W_\infty(\bar\mu^{N_2}, \mu_k) + W_\infty(\bar\mu^{N_2}, \mu) \leq 5 \varepsilon$, which yields the claim.
Best Answer
The set $MS=\{(\mu,K)\in P_X\times K_X:\mathrm{supp}(\mu)=K\}$ is of type $G_\delta$ in $P_X\times K_X$ and hence the weak+Hausdorff topology on $P_X$ is Polish.
Indeed, fix any countable base $\{U_n\}_{n\in\mathbb N}$ of the topology of $X$ and observe that $$MS=\bigcap_{n\in\mathbb N}\{(\mu, K)\in P_X\times K_X:\mu(U_n)>0\;\Leftrightarrow\;U_n\cap K\ne\emptyset\}.$$ So, it remains to show that for every $n\in\mathbb N$ the set $$MS_n:=\{(\mu, K)\in P_X\times K_X:\mu(U_n)>0\;\Leftrightarrow\;U_n\cap K\ne\emptyset\}$$is of type $G_\delta$ in $P_X\times K_X$.
Observe that $(P_X\times K_X)\setminus MS_n=MS_n'\cup MS_n''$ where $$MS_n':=\{(\mu,K)\in P_X\times K_X: \mu(U_n)>0\;\wedge\;U_n\cap K=\emptyset\}$$ and $$MS_n'':=\{(\mu,K)\in P_X\times K_X: \mu(U_n)=0\;\wedge\;U_n\cap K\ne\emptyset\}.$$ The known properties of the topologies on the spaces $P_X$ and $K_X$ ensure that the set $\{\mu\in P_X:\mu(U_n)>0\}$ is open in $P_X$ and the set $\{K\in K_X:K\cap U_n\ne \emptyset\}$ is open in $K_X$. Then $MS_n'$ and $M_s''$ are intersections of open and closed sets, so are of type $F_\sigma$ in $P_X\times K_X$. Then $MS_n$ is of type $G_\delta$ in $P_X\times K_X$ and so is the set $MS=\bigcap_{n\in\mathbb N}MS_n$.