If $E$ is a locally compact metric space and $f:[0,∞)→E$ is càdlàg, is $f$ càdlàg as a function into the one-point compactification of $E$ too? Let $(E,\tau)$ be a locally compact Hausdorff space, $\infty\not\in E$ be an abstract point, $E^\ast:=E\uplus\left\{\infty\right\}$ and $\tau^\ast:=\tau\cup\left\{E\setminus K\cup\left\{\infty\right\}:K\subseteq E\text{ is compact}\right\}$. $(E^\ast,\tau^\ast)$ is a compact Hausdorff space known as the Alexandroff one-point compactification of $(E,\tau)$.
Assume $\tau$ is generated by a metric $d$ on $E$.

Question 1: Are we able to infer that $\tau^\ast$ is generated by a metric $d^\ast$ on E$^\ast$? If so, how are $d$ and $d^\ast$ related? (I've read that $(E^\ast,\tau^\ast)$ is always metrizable when $(E,\tau)$ is separable; even when $\tau$ is not generated by a metric. However, if necessary, assume that $(E,\tau)$ is separable.)
Question 2: If $f:[0,\infty)\to E$ is càdlàg, is $f$ still càdlàg when considered as function into $E^\ast$?

 A: Question 1: The Alexandroff compactification of a metrizable space is metrizable if and only if the original space is separable. Compact metrisable spaces are easily seen to be separable via the following lemma.

Lemma: If $X$ is a totally bounded metric space then $X$ is separable.
Proof: For each $n$ there is a finite $\frac1n$-net $A_n$ for $X$ (so for all $x \in X, d(x,A_n) < \frac1n$). Then $\bigcup_{n \geq 1} A_n$ is a countable dense subset of $X$.

In particular, if $E^*$ is metrisable, $E$ is homeomorphically embedded in a separable space and hence is separable.
The converse follows from Urysohn's metrization theorem. It's obvious that $E^*$ is regular Hausdorff since it is compact and Hausdorff. It is a bit more fiddly but not very difficult to show that it is second countable by using the fact that $E$ itself must be second countable since it is a separable metric space.
However the converse obviously fails if $E$ is not metrisable since subspaces of metrisable spaces are metrisable.
Question 2: Part of the construction of $E^*$ includes that $E$ is homeomorphically embedded in to $E^*$ via some map $i$. It is then immediate that $f$ is càdlàg as a function valued in $E^*$ since the topological definition of limits is suitably compatible with the definition of the subspace topology.
For example, you want to check that left limits of $f$ exist in $E^*$ and are the same as the limits in $E$.
This means you want to show that for every $t \in (0,\infty)$ and every open neighbourhood $O^*$ of $\lim_{s \uparrow t} f(s)$ in $E^*$ (where the limit is taken in $E$) there is an $\varepsilon >0$ such that $f(s-\varepsilon,s) \subset O^*$. But $O^* \cap E$ is an open neighbourhood of $\lim_{s \uparrow t} f(s)$ in $E$ and so there is an $\varepsilon$ such that $f(s-\varepsilon,s) \subset O^* \cap E$ since is cadlag as an $E$-valued function.
The right-continuity is then similar.
