Solution:
(Ⅰ) From the given conditions, we have $\begin{cases}2a=2\sqrt{3} \\ \frac{c}{a}= \frac{\sqrt{3}}{3}\end{cases}$, solving these equations gives $a=\sqrt{3}$, $c=1$. Therefore, $b^2=a^2-c^2=2$.
Thus, the equation of the ellipse $C$ is $\frac{x^2}{3}+ \frac{y^2}{2} =1$.

(Ⅱ) Proof: According to the problem, the equation of circle $O$ is $x^2+y^2=3$.
Let $N(3,t)$, $M(x_0,y_0)$, where $x_0^2 +y_0^2 =3$.
From $|ON|^2=3+|MN|^2$, we get $3^2+t^2=3+(x_0−3)^2+(y_0−t)^2$,
which simplifies to $9+t^2=3+x_0^2 −6x_0+9+y_0^2 −2ty_0+t^2$,
thus $3+x_0^2 −6x_0+y_0^2 −2ty_0=0$.
Since $x_0^2 +y_0^2 =3$, we have $3x_0+y_0t-3=0$.
When $t=0$, $x_0=1$, the equation of line $l$ is $x=1$, and line $l$ passes through the right focus of ellipse $C$, $F(1,0)$.
When $t\neq 0$, the equation of line $MN$ is $y−y_0=-\frac{3}{t} (x−x_0)$,
which can be rewritten as $ty-ty_0=-3x+3x_0$, thus $ty=-3(x-1)$, and line $l$ passes through the right focus of ellipse $C$, $F(1,0)$.
In conclusion, line $l$ passes through the right focus of ellipse $C$, $F(1,0)$.

Therefore, the equation of the ellipse $C$ is $\boxed{\frac{x^2}{3}+ \frac{y^2}{2} =1}$, and it is proven that the line passing through point $M$ and perpendicular to line $ON$ passes through the right focus of the ellipse $C$, $\boxed{F(1,0)}$.