$\newcommand{\de}{\delta}
\newcommand{\vp}{\varepsilon}$

Take any $x,y$ in $D:=D[0,1]$ with $d(x,y)\le\vp$ for some $\vp\in(0,1)$, so that for some strictly increasing continuous function $\ell$ mapping $[0,1]$ onto itself we have $\|y\circ\ell-x\|\le\vp$ and $\|\ell-\text{id}\|\le\vp$, where $\|\cdot\|:=\|\cdot\|_\infty$.

Since $x\in D$, it is easy to see that $\|x\|<\infty$. So, $\|y\|=\|y\circ\ell\|\le\|x\|+\vp<\infty$.

For $t\in[0,1]$, let
\begin{equation}
y^+(t):=\sup_{0\le s\le t}y(s)
\end{equation}
and $y^-:=y^+-y$. Then $y^\pm$ is a nondecreasing function in $D$. So, there exists a unique nonnegative (Lebesgue--Stieltjes) measure $\mu^\pm$ such that $\mu^\pm([0,t])=y^\pm(t)-y(0)$ for all $t\in[0,1]$. Moreover, $\|\mu^+\|_{tv}=y^+(1)-y(0)\le2\|y\|$ and similarly $\|\mu^-\|_{tv}\le2\|y\|$, where $\|\cdot\|_{tv}$ denotes the total variation norm. Letting now
\begin{equation}
\mu:=\mu^0+\mu^+-\mu^-,
\end{equation}
where $\mu^0:=y(0)\de_0$ and $\de_0$ is the Dirac measure supported at $0$, we have $\|\mu\|_{tv}\le5\|y\|$ and
\begin{equation}
y(t)=\int_{[0,t]}d\mu=\int I_{[0,t]}d\mu
\end{equation}
for $t\in[0,1]$, where $\int:=\int_{[0,1]}$ and $I$ denotes the indicator.

Hence, for the $L^2$ norm $\|\cdot\|_2$ we have
\begin{align}
\|y\circ\ell-y\|_2^2&=\int dt\,(y(\ell(t))-y(t))^2 \\
&=\int dt\Big(\int\mu(du)\,\big(I_{[0,\ell(t)]}(u)-I_{[0,t]}(u)\big)\Big)^2 \\
&=\iint \mu(du)\mu(dv)\,\int dt\, J(u,v,t),
\end{align}
where
\begin{equation}
J(u,v,t):=(I_{[0,\ell(t)]}(u)-I_{[0,t]}(u)\big)\big(I_{[0,\ell(t)]}(v)-I_{[0,t]}(v)\big),
\end{equation}
so that
\begin{align}
\big|J(u,v,t)\big|
&\le\big|I_{[0,\ell(t)]}(u)-I_{[0,t]}(u)\big|
\\
&=I\{u\text{ btw }\ell(t)\text{ and }t\} \\
&= I\{t\text{ btw }\ell^{-1}(u)\text{ and }u\} \\
\end{align}
where btw means "is between". So,
\begin{equation}
\int dt\, |J(u,v,t)|\le\max_u|\ell^{-1}(u)-u|=\max_t|\ell(t)-t|\le\vp
\end{equation}
and hence
\begin{equation}
\|y\circ\ell-y\|_2^2\le\|\mu\|_{tv}^2\,\vp\le(5\|y\|)^2\vp
\le25(\|x\|+\vp)^2\vp.
\end{equation}
Also,
\begin{equation}
\|y\circ\ell-x\|_2\le\|y\circ\ell-x\|\le\vp.
\end{equation}
Thus,
\begin{equation}
\|y-x\|_2\le \|y\circ\ell-x\|_2+\|y\circ\ell-y\|_2\le\vp+5(\|x\|+\vp)\sqrt{\vp}
\le6(\vp+\|x\|\sqrt{\vp}\,),
\end{equation}
which proves the desired continuity.

One may also note that the latter bound is optimal up to a universal constant factor. Indeed, let $\vp\in(0,\frac12)$, $x=cI_{[\frac12,1]}$, and $y=\vp+cI_{[\frac12+\vp,1]}$, where $c$ is any real number. Then $x$ and $y$ are in $D$, and $d(x,y)\le\vp$ (consider the function $\ell$ that is affine on each of the intervals $[0,\frac12]$ and $[\frac12,1]$ and maps $0,\frac12,1$ to $0,\frac12+\vp,1$, respectively). On the other hand, it is easy to see that here $\|y-x\|_2\asymp\vp+|c|\sqrt{\vp}\asymp\vp+\|x\|\sqrt{\vp}$ -- which proves the mentioned optimality.