Another way of solving this problem is with bounding boxes. A bounding box is the minimum rectangle containing a shape orthogonal to two chosen axes. Consider the bounding boxes of the two squares A and B orthogonal to the main square Q. In each case, the remaining space makes four right angled triangles. Call them the waste triangles.
Now you need two lemmas.
Given two non-overlapping squares A' and B', A' either doesn't intersect the bounding box of B', or it does so only in one of the waste triangles. Proof: assume that there is an intersection in more than one triangle. Draw a line between the two points in the two distinct triangles. This line intersects B' at some point. But by convexity of the square, this point must also be in A'. Which is impossible. And we're done.
A square can always be placed in the corner of its bounding box so that it does not intersect the triangle at the opposite corner. For this, use algebra and express everything in terms of the two non-hypotenuse sides of the waste triangle. The bottom left corner of the bounding box is the origin, so all squares in this corner have their top-right corner along the line x = y. The intersection of this line and the line formed by the hypotenuse of the top-right waste triangle gives a value of x that is bigger than the side of the square. And so we're done.
Now we're ready for the main proof. Take the square A. We can always put it into some corner of its bounding box by our two lemmas. Do this. Repeat the process for B. Now we have two squares orthogonal to the main square. Showing that the sides add up to at most 1 is now trivial.
Proof of Lemma 2: We look for the intersection of the diagonal $x = y$ with the line extended from the hypotenuse of the top-right triangle. That is, the green and red lines from the image below:

Now the equation of the red line is $y = -(\frac{a}{b})x + c$ for some $c$.
To find this c, we just use the fact that that value of $y$ at $x = a$ for the red line is $a + b$. Putting this into the equation we get: $ a + b = -(\frac{a^2}{b}) + c$. From here we get $ c = (\frac{a^2 + b^2 + ab}{b})$. So the equation of our line becomes: $y = -(\frac{a}{b})x + \frac{a^2 + b^2 + ab}{b}$. Intersecting with the line $x = y$, the green line, we get: $x = -(\frac{a}{b})x + \frac{a^2 + b^2 + ab}{b}$, and after some algebraic manipulation, this works out as: $x = \frac{a^2 + b^2 + ab}{a + b}$.
At this point, it's good to reiterate the meaning of $x$. This is the size of the largest square that fits orthogonally into the bottom left corner of the black bounding box without intersecting the top-right waste triangle. Equivalently, it is the largest such square that doesn't intersect the red line. All we have now to prove, to show that the blue square can be placed in this corner, is that the side of the blue square is less than or equal $x$. So our goal is to solve the inequality:
$\sqrt{a^2 + b^2} \leq x$
Substitute our term for $x$:
$\sqrt{a^2 + b^2} \leq \frac{a^2 + b^2 + ab}{a + b}$
Positive terms allows squaring both sides:
$a^2 + b^2 \leq \frac{(a^2 + b^2 + ab)^2}{(a + b)^2} \equiv$ $(a^2 + b^2)(a + b)^2 \leq (a^2 + b^2 + ab)^2$
Now let $u = a^2 + b^2$ yielding:
$u(a + b)^2 \leq (u + ab)^2$
Now multiply it all out:
$ua^2 + ub^2 + 2uab \leq u^2 + a^2b^2 + 2uab \equiv$ $ua^2 + ub^2 \leq u^2 + a^2b^2$
And plug back in our term for $u$ getting:
$a^4 + 2a^2b^2 + b^4 \leq a^4 + b^4 + 2a^2b^2 + a^2b^2 \equiv$ $0 \leq a^2b^2$
Which is clearly true. And we're done proving the lemma.