Since the squares are convex, we can draw lines which separate them. In particular, if two separating lines go from $(b-a,0)$ to $(b,1)$ and from $(c,1)$ to $(c+d,0)$, then we can prove the result in terms of those lines and those variables.
So: let the rectangle go from $(0,0)$ to $(2,1)$. Let $A$ be the leftmost square (or one such square). Let $C$ be the rightmost square (or one such square). Let $B$ be the other square.
Draw a line separating $A$ and $B$, and let $(b,1)$ and $(b-a,0)$ be its intersections with the lines $y=1$ and $y=0$.
Draw a line separating $B$ and $C$, and let $(c,1)$ and $(c+d,0)$ be its intersections with the lines $y=1$ and $y=0$.
Reasoning as in user21820's answer, we assume wlog that:
- $0<a$, so $A$ is left and above the line separating $A$ and $B$;
- $0<d$, so $C$ is right and above the line separating $B$ and $C$;
- $0<b$ and $c<2$ and $a-b+c+d>0$, so $B$ is below both lines.
(Since the lines may leave the rectangle, we do not assume $b-a>0$ or $b<2$ or $c>0$ or $c+d<2$.)
Lemma (proved at the end):
\begin{align}
\text{sidelength of }A &\le \min\!\left(\frac{b}{a+1},\,1\right)\\
\text{sidelength of }B &\le \min\!\left((a-b+c+d)u,\,1\right)\\
\text{sidelength of }C &\le \min\!\left(\frac{2-c}{d+1},\,1\right)
\end{align}
where
$$u=\max\left(
\frac{1}{a+d+1},
\frac{\sqrt{a^2+1}}{a^2+a+d+1},
\frac{\sqrt{d^2+1}}{d^2+d+a+1}
\right)$$
The factor $u$ satisfies $1/(a+1)>u$ and $1/(d+1)>u$ so long as $a<3.66$ and $d<3.66$ respectively. I will assume those inequalities for now to show that some functions are increasing or decreasing; I don't have a clean proof for those inequalities or without them yet.
In the corner case of $b=a+1$ and $c=1-d$, the side lengths are $1$, $0$ and $1$, and they sum to exactly $2$. We now use this in analyzing four cases.
Case I, $b\le a+1$ and $c\le 1-d$: The sum of the sidelengths is at most
$$\frac{b}{a+1}+(a-b+c+d)u+1$$ This is increasing in both $b$ and $c$, so its value is at most the corner value of $2$.
Case II, $b\le a+1$ and $c\ge 1-d$: The sum of the sidelengths is at most
$$\frac{b}{a+1}+(a-b+c+d)u+\frac{2-c}{d+1}$$ This is increasing in $b$ and decreasing in $c$, so its value is at most the corner value of $2$.
Case III, $b\ge a+1$ and $c\le 1-d$: The sum of the sidelengths is at most
$$1+(a-b+c+d)u+1$$
This is decreasing in $b$ and increasing in $c$, so its value is at most the corner value of $2$.
Case IV, $b\ge a+1$ and $c\ge 1-d$: The sum of the sidelengths is at most
$$1+(a-b+c+d)u+\frac{2-c}{d+1}$$
This is decreasing in both $b$ and $c$, so its value is at most the corner value of $2$.
So the sum of the sidelengths is at most $2$ in each case.
Proof of Lemma:
The sidelength of $A$ is clearly less than 1, and also clearly less than the maximum sidelength inscribed in the right triangle bounded by $x=0$, $y=1$, and the separator of $A$ and $B$. We use Polya's formula here to calculate the sidelength in the triangle as the maximum of $b/(a+1)$ and $b\sqrt{a^2+1}/(a^2+a+1)$; since $a>0$, the maximum is just $b/(a+1)$.
A similar use of Polya's result bounds the sidelength of $C$. Yet another use of that result, now for a triangle which may be acute or obtuse, bounds the sidelength of $B$ by $|a-b+c+d|u$. Since we assumed $a-b+c+d>0$, we write this bound as $(a-b+c+d)u$.
