A simple real analysis appoach is sufficient for this.However, I think a more graph-like approach is much better for your develop.
Image $f$ as a curve $y=f(x)$ , that curve is in $Oxy$ and must sasitfy these:
i) $f$ connects 2 point in line $x=0$ and $x=1$ respectively
ii)$f$ lies entirely in the cell $[0;1]\times[0;1]$
iii) (Not necessary for this problem) each line $x=c(0=c=1)$ cuts $f$ at exactly 1 point.
Thus, the question of problem is to find which $\alpha$ so that $y=\alpha x$ and $f$ intersect forall all curve $f$ sastify i) and ii)
(Try to draw yourself some lines and curves , to know clearlier )
We find that the line $y=\alpha x$ devides the cell in two parts .
Thus, the question of problem occurs if and only if in that cell,line $x=1$ and $x=0$ lies in two different parts( not include endpoints) which are devided by y=\alpha x
(Again, try draw some, it's good for anyone's knowledge)
Which equivalent to $\alpha \ge 1 $ .
In case, you want an alternative proof:
For $\alpha <1$ , we choose the function $f_1(x)=1$, thus we easily see that $f_1(x)>\alpha alpha \forall x \in [0;1]$.Hence, these $\alpha$ do not sasity the conclusion.
For $\alpha \ge 1$ , $f(x) \in [0;1]$, therefore $ [ f(0)-\alpha.0][f( 1) -\alpha.1] \le 0$.
Due to the continuousness of $ f(x)-\alpha x$, we imply the conclusion.