Suppose we have a sequence of $\mathbb{R}^d$-valued stochastic processes $(X^n)_{n \in \mathbb{N}}$ converging to $X$ uniformly on compacts in probability, i.e. $$ \operatorname{plim}_{n \to \infty} \sup\limits_{s \in [0,N]} ||X_s^n - X_s||_{\mathbb{R}^d} = 0 \;\; \forall N \in \mathbb{N}, $$ where plim denotes convergence in probability. Equivalently we could write that for all $t \geq 0, \varepsilon > 0$ $$ \mathbb{P}[\sup_{s\leq t}||X^n_s-X_s||_{\mathbb{R}^d}>\varepsilon]\to 0. $$
I now want to know whether the ucp-convergence in probability is preserved under continuous transformations, i.e. if $f: \mathbb{R}^d \to \mathbb{R}^k$ is continuous, then $f(X^n)$ converges ucp to $f(X)$ in probability.
I have the intuition that this statements holds (since we are looking at compact sets, the function is even uniformly continuous and hence doesn't explode too much), but I am not able to show it. I have already thought about the more specific case of Lipschitz-continuity. For this type of continuity I can show that the statement holds, because for $N \in \mathbb{N}$ it holds that $$ \sup_{s \in [0,N]}||f(X^n_s)-f(X_s)||_{\mathbb{R}^k} \leq L \sup_{s \in [0,N]}||X^n_s-X_s||_{\mathbb{R}^d} \to L *0 = 0, $$ where $L>0$ is $f$'s Lipschitz-constant, hence the limit in propability can be estimated from above by $0$. Turning to the more general case, I thought about applying the $\varepsilon$-$\delta$-criteria. Since I am looking at compact intervals of the form $[0,N]$, I know that $f$ is uniformly continuous there. Hence given a $\varepsilon > 0$ there exists a $\delta_\varepsilon > 0$ such that $$ || X^n_s - X_s ||_{\mathbb{R}^d} < \delta_\varepsilon \implies || f(X^n_s) - f(X_s) ||_{\mathbb{R}^k} < \varepsilon $$ for $s \in [0,N], N \in \mathbb{N}$. Now I don't know hot to go on. Does the statement even hold? If no, is there a nice counterexample?