Let $w:[0,1]\rightarrow [0,1]$ be an increasing function with $w(0)=0,w(1)=1$ and such that: \begin{align} w(p)+w(q)<w(p+q)\quad \forall p,q \quad st \quad 0<p+q\leq 1 \quad\&\quad 0<p,q<1 \end{align} I am asked to prove that $w(p)<p$ $\forall p\in(0,1)$.
After some manipulations I have come to the following expression: \begin{align*} w(\lambda r)+w((1-\lambda) r)<w(r) \quad \forall \lambda \in(0,1),\forall r\in(0,1] \end{align*} and I would like to prove that $w(\lambda r)<\lambda w(r)$ as this would solve the problem. However, I do not seem to find any proof for this. Any hints?
What you are trying to prove is not true: I will give an example of a function $w\colon [0,1]\to[0,1]$ that satisfies all given conditions, but does not satisfy $w(p)<p$ for all $p\in (0,1)$.
Consider the following function $$ f(x)=\begin{cases}\tfrac{x^2}{2},&x\in[0,\tfrac12]\\\frac{x^2+1}{2},&x\in(\tfrac12,1]\end{cases} $$ Then $f(x)$ satisfies the following conditions. It is increasing (with a jump discontinuity at $x=\tfrac12$). It satisfies $f(0)=0$ and $f(1)=1$. Most importantly, it satisfies $f(x)+f(y)<f(x+y)$ for all $x,y\in(0,1)$ such that $x+y\leq 1$.
To verify the claim I just made, it suffices to consider two cases: either $x+y>\tfrac12$, or not. In the former case, we must show that $$ \frac{x^2+y^2+n}{2}<\frac{(x+y)^2+1}{2}, $$ where $n$ is the number of elements of $\{x,y\}$ that are strictly greater than $\tfrac12$. Since $x+y\leq 1$, we must have $n<2$, so that $$ f(x)+f(y)=\frac{x^2+y^2+n}{2}\leq \frac{x^2+y^2+1}{2}<xy+\frac{x^2+y^2+1}{2}=\frac{(x+y)^2+1}{2}=f(x+y), $$ proving the claim in the case $x+y>\tfrac12$.
In the remaining case $x+y\leq \tfrac12$, everything is simpler and we have $$ f(x)+f(y)=\frac{x^2+y^2}{2}<xy+\frac{x^2+y^2}{2}=f(x+y). $$
So $f$ satisfies all the conditions given. However, it does not satisfy $f(x)<x$ for all $x\in(0,1)$. An explicit counterexample is obtained at $x=\tfrac34$ for instance, since $$ f\bigl(\tfrac34\bigr)=\frac{25}{32}>\frac{24}{32}=\frac{3}{4}. $$