Let $C([0,T], \mathbb{R}^d)$ be the metric space of continuous functions from $[0,T]$ to $\mathbb{R}^d$, endowed with the supremum metric. For any metric space $E$, let $\mathcal{P}(E)$ be the space of probability measures on $E$, endowed with the weak topology.
For fixed $t \in [0,T]$, we define the map
$$ \Phi: \mathcal{P}(C([0,T],\mathbb{R}^d)) \to \mathcal{P}(\mathbb{R}^d); \quad \quad \mu \mapsto \mu_t,$$ where $\mu_t$ is the marginal of $\mu$ at time $t$.
Is it true that $\Phi$ is continuous? It seems obvious, but I can't see that from the definition.
A formal proof proceeds as the following. We denote by $\omega$ an element of the sample space $C([0,T])$ and by $\omega(t)$ its $t$-coordinate. By definition, we have $$ \mu_t(A)=\mu\left(\left\{\omega\in C([0,T]):\omega(t)\in A\right\}\right)=\int 1_{\omega(t)\in A}\mathrm d\mu $$ for any fixed $t\in [0,T]$ and $A\in\mathcal{B}(\Bbb R^d)$. This implies $$ \int_{\Bbb R^d} 1_A\mathrm{d}\mu_t = \int_{C([0,T])} 1_A(\omega(t))\mathrm{d}\mu(\omega) $$ for all $A\in\mathcal{B}(\Bbb R^d)$, and thus $$ \int_{\Bbb R^d} f\;\mathrm{d}\mu_t=\int_{C([0,T])} f(\omega(t))\;\mathrm{d}\mu(\omega) $$ holds for all bounded continuous $f:\Bbb R^d\to \Bbb R$.
To prove weak continuity of the map $\Phi_t:\mu\mapsto\mu_t$, we need to show $\Phi_t(\mu_n)\to \Phi_t(\mu)$ weakly whenever $\mu_n\to \mu$ weakly. Assume $\mu_n$ converges to $\mu$ weakly, i.e. for every bounded continuous functional $F:C([0,T])\to\Bbb R$, it holds $$ \int_{C([0,T])}F(\omega)\mathrm{d}\mu_n(\omega)\to \int_{C([0,T])}F(\omega)\mathrm{d}\mu(\omega).\tag{*} $$ Observe that $\omega \mapsto f(\omega(t))$ is a bounded continuous functional on $C([0,T])$ for any bounded continuous function $f:\Bbb R^d \to \Bbb R$. This is obvious since if $\omega_n(\cdot) \to \omega(\cdot)$ uniformly on $[0,T]$, then $\omega_n(t)\to\omega(t)$ and $f(\omega_n(t))\to f(\omega(t))$ for each $t\in[0,T]$. Thus, $(*)$ implies $$ \int_{\Bbb R^d}f\mathrm{d}\Phi_t(\mu_{n})=\int_{C([0,T])}f(\omega(t))\mathrm{d}\mu_n(\omega)\to \int_{C([0,T])}f(\omega(t))\mathrm{d}\mu(\omega)=\int_{\Bbb R^d}f\mathrm{d}\Phi_t(\mu) $$ for all bounded continuous $f$. This proves the claim.