### Solution:

#### Part (1) Proof:
- Given: The solution set of the equation $x^{2}+a=x$ where $a\in \mathbb{R}$ is denoted as $A$ ($A\neq \varnothing$).
- Let $x_{0}\in A$, so we have the equation ${x_0}^2+a={x_0}$.
- Now, consider the equation $(x^{2}+a)^{2}+a=x$. We substitute $x_{0}$ for $x$ in this equation, which becomes:
$$((x_0^{2}+a)^{2}+a=x_0)$$
Given that $x_0^{2}+a=x_0$, this substitution holds true, meaning $x_{0}$ is also a solution of $(x^{2}+a)^{2}+a=x$.
- Therefore, $x_{0}\in B$. Since this is true for any $x_0 \in A$, we conclude $A\subseteq B$.

#### Part (2) Finding the Range of $a$ for $A=B$:
- Knowing $A\neq \varnothing$, the equation $x^{2}-x+a=0$ must have real roots, leading to the discriminant condition $\Delta =1-4a\geqslant 0$, simplifying to $a\leq \frac{1}{4}$.
- For $B$ to equal $A$, the polynomial $x^{4}+2ax^{2}-x+a^{2}+a=0$ must factorize to include $(x^{2}-x+a)$, resulting in $(x^{2}-x+a)(x^{2}+x+a+1)=0$.
- For $x^{2}+x+a+1=0$ to have no real roots, the discriminant must be negative: $1-4(a+1) \lt 0$, which simplifies to $a \gt -\frac{3}{4}$.
- For the roots of $x^{2}+x+a+1=0$ to match those of $x^{2}-x+a=0$:
- Distinct real roots are not possible per Vieta's formulas since the sums and products of the roots would not match.
- Equal real roots occur when $\Delta =1-4(a+1)=0$, giving $a=-\frac{3}{4}$. The root $x=-\frac{1}{2}$ satisfies both equations.
- Combining these conditions, the range for $a$ when $A=B$ is $\boxed{\{a|-\frac{3}{4}\leq a\leq \frac{1}{4}\}}$.