Given that you're asking about "expectation" rather than "integral" operators and talking about the Levy-Prokhorov metric, I'm going to assume that you meant that $\Delta[0, 1]$ is the set of probability measures on the Boral $\sigma$-algebra.
In this case, yes, the expectation operator is an open map.
I'll be using the coupling representation of the Prokhorov metric mentioned in this answer: https://math.stackexchange.com/a/4310676/154826
That is,
$$
d_P(\mu, \nu)
= \inf\{\alpha(X, Y) : X \sim \mu, Y\sim \nu\},
$$
where the infimum is taken over all couplings $(X, Y)$, and where
$$
\alpha(X, Y)
= \inf\{\epsilon : \Pr(|X - Y| > \epsilon) \leq \epsilon\}.
$$
We also define the expectation operator $E \colon \Delta[0, 1] \rightarrow [0, 1]$ by $E(\mathcal{L}_X) = \mathbf{E} X$, where $\mathcal{L}_X$ is the law of $X$.
Since all probability measures on $[0, 1]$ can be expressed as laws of random variables, this is well-defined.
So now, in order to demonstrate that $E$ is an open map, we must show that for any open ball $B_\epsilon(\mathcal{L}_X)$, the image $E(B_\epsilon(\mathcal{L}_X))$ contains an open neighbourhood of $E(\mathcal{L}_X) = \mathbf{E} X$.
When $\mathbf{E} X \in (0, 1)$, we will find an open interval
$$
I =\{\kappa \mathbf{E}X : \kappa \in (\kappa_-, \kappa_+)\},
$$
where $\kappa_- < 1 < \kappa_+$ such that $I \subseteq E(B_\epsilon(\mathcal{L}_X))$.
That is, for any $\kappa \mathbf{E} X \in I$, we will show that there is a coupling $(X, Y)$ such that $E(\mathcal{L}_Y) = \mathbf{E} Y = \kappa \mathbf{E} X$ and
$$
\Pr(|X - Y| < \epsilon/2) \leq \epsilon/2
$$
so that $d_P(\mathcal{L}_X, \mathcal{L}_Y) < \epsilon$ and so $\mathcal{L}_Y \in B_\epsilon(\mathcal{L}_X)$.
So let $\kappa \mathbf{E}X \in I$.
If $\kappa <1$, define
$$
Y = \kappa X.
$$
This takes values in $[0, \kappa] \subseteq [0, 1]$, and $\mathbf{E} Y = \kappa \mathbf{E}X$.
Moreover, we have that
\begin{align*}
\Pr(|X - Y| > \epsilon/2)
= \Pr((1 - \kappa) X >\epsilon/2)
= \Pr\biggl(X >\frac{\epsilon/2}{1 - \kappa}\biggr),
\end{align*}
which is $0$ when $\kappa \geq \kappa_- := 1 - 2\epsilon$.
On the other hand, consider the case $\kappa > 1$.
In this case, we define
$$
Y = \lambda + (1 - \lambda)X,
$$
where
$$
\lambda
= \frac{1 - \kappa \mathbf{E} X}{1 - \mathbf{E} X}
= 1 + (1 - \kappa) \frac{\mathbf{E} X}{1 - \mathbf{E} X}.
$$
You can check that this is the linear function that takes $1\mapsto 1$ and $\mathbf{E} X \mapsto \kappa \mathbf{E}X$, and so in particular that $\mathbf{E}Y = \kappa \mathbf{E} X$.
Moreover, we can see that, since $\kappa > 1$, we have that $\lambda \in [0, 1]$, and so $Y$ takes values in $[\lambda, 1] \subseteq [0, 1]$.
Last, we compute
\begin{align*}
\Pr(|X - Y| > \epsilon/2)
= \Pr((1 - \lambda)(1 - X) > \epsilon/2)
= \Pr\biggl(1 - X > \frac{\epsilon/2}{1 - \lambda}\biggr),
\end{align*}
which is $0$ when $1 - \lambda < \epsilon/2$.
But since $\lambda$ is a decreasing function of $\kappa$, this is equivalent to $\kappa < \kappa_+$ for some $\kappa_+ > 1$.
Therefore, with these $\kappa_- < 1 < \kappa_+$, we have that for any $\kappa \in (\kappa_-, \kappa_+)$, there exists $\mathcal{L}_Y \in B_{\epsilon}(\mathcal{L}_X)$ such that $\mathbf{E} Y = \kappa \mathbf{E} X$ and so $I \subseteq E(B_{\epsilon}(\mathcal{L}_X))$.
The cases $\mathbf{E} X \in \{0, 1\}$ are easier, so I leave them to you.
Thus, we've shown that, for any probability measure $\mathcal{L}_X \in \Delta[0, 1]$, the image of any $E(B_\epsilon(\mathcal{L}_X))$ contains an open neighbourhood of $\mathbf{E} X$, and so that $E$ is an open map.
