To reach maximum area we have to put one of vertices of triangle as a vertice of square. Otherwise we will have smaller value of height or base. Put square on a Cartesian plane as $(0,0),(1,0),(0,1),(1,1)$. Let $a=(1,0), b=(0,y), c=(x,1)$ be vertices of triangle where $x,y\in [0;1]$. By the formula of triangle $S=1/2|(x_b-x_a)(y_c-y_a)-(y_b-y_a)(x_c-x_a)|=1/2|-1-y(x-1)|=1/2|-1+y(-x+1)|$ We have $0\le y\le 1\wedge 0\le 1-x\le 1\implies 0\le (1-x)y\le 1 \iff -1\le -1+y(-x+1)\le 0\implies |-1+y(-x+1)|\le 1$ hence $S\le 1/2$ with equality iff $x=1$ or $y=0$. So maximum attained iff one of the side of triangle is a side of square and the third vertices lies on the opposite side of square.

Please me more concise and clear!

Justification is above. The final answer is $\frac{1}{2}$. I don't see anything unclear.

Here is my old prove: https://imgur.com/a/rCaDA

I like this task a lot. It's such a classical & essential olympic task. Maybe easy, but still out of reach for most of A-graded students