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.
