The closest point of an ellipse

Squared distance.
$$D(x)=(x-1)^2+1-\frac{x^2}4=\frac34x^2-2x+2,\qquad x\in[-2,2].$$

Minimum. $D'(x)=\frac32x-2=0\iff x=\frac43\in[-2,2]$. The parabola opens upward so this is the global minimum on $[-2,2]$:
$$D\left(\frac43\right)=\frac34\cdot\frac{16}9-\frac83+2=\frac23.$$

Answer. $x=\frac43$, $y=\pm\sqrt{1-\frac49}=\pm\frac{\sqrt5}3$, and the distance is $\sqrt{\dfrac23}\approx0.816$.

Check. The vertex $(2,0)$ is at distance $1>0.816$, so the closest point is indeed not the nearest vertex.

Remark. At a closest point, the segment from $F$ to the point is perpendicular to the tangent. Here the tangent at $\left(\frac43,\frac{\sqrt5}3\right)$ has slope $-\frac{x}{4y}=-\frac1{\sqrt5}$ and the segment has slope $\frac{\sqrt5/3}{1/3}=\sqrt5$: indeed $-\frac1{\sqrt5}\cdot\sqrt5=-1$.