How can I prove $\log_2{(1+x)} \geq {x}$ for $0<x<1$

The logarithm function is concave.

Thus, it lies above any secant line.

In particular, on $[0,1]$, the function $x$ is a secant line for $\log_2(1+x)$.

Hence, $\log_2(1+x)\ge x$ on $[0,1]$.

Hint: Exponentiate the inequality to the equivalent form $1+x\ge2^x$, and sketch the two graphs $y=1+x$ and $y=2^x$. Note that $(x,y)=(0,1)$ and $(1,2)$ are common to both graphs.