I'm reading Billingsley's "Convergence of probability measures" (1968), p. 111. The definitions are: $D$ - the space of $\textit{cadlag}$ functions on [0,1], $\Lambda$ - the class of strictly increasing, continuous mappings of [0,1] onto itself. For $x,y\in D$ define $d(x,y):=\inf\{\varepsilon>0:\ \exists\lambda\in\Lambda:\ \sup_t|\lambda(t)-t|<\varepsilon \text{ and } \\ \sup_t|x(t)-y(\lambda(t))|<\varepsilon\}.\tag{1}$ I'm stuck with the proof that $d(x,y)=0$ implies $x=y$. The author claims: "$d(x,y)=0$ implies that for each $t$ either $x(t)=y(t)$ or $x(t)=y(t-)$, which in turn implies $x=y$."
This is what I tried: let $x,y\in D$, $d(x,y)=0$ and $t\in[0,1]$. Now from (1) it follows that there exists a sequence $(z_n)\subset[0,1]$ such that $z_n\to t$ and $y(z_n)\to x(t)$ as $n\to\infty$. If $t$ is a continuity point of $y$, then $y(z_n)\to y(t)$, thus $x(t)=y(t)$. If $y$ has a jump at $t$ and $(z_n)$ has a subsequence $(z_{n_k})$ such that $z_{n_k}\geq t, \forall k$, then $y(z_{n_k})\to x(t)$ and $y(z_{n_k})\to y(t)$, as $k\to\infty$, thus $y(t)=x(t)$. Otherwise, $z_n
Now suppose that $y=\mathbf{1}_{[t,1]}$, then $y$ has a jump at $t$ and if $d(x,y)=0$, then $x=\mathbf{1}_{(t,1]}$. But this is a contradiction since $x\neq y$ and $x$ is not $\textit{cadlag}$.
Where am I wrong?