Consider the Taylor series for $\log(1+x)$: $\log(1+x) = x - \frac{x^2}{2}+\frac{x^3}{3} - \frac{x^4}{4}+\cdots + (-1)^{n+1}\frac{x^n}{n}+\cdots$ which holds for $|x|\lt 1$ and $x=1$. So on $[0,1]$, this can be used to show $\log(1+x)\leq x$ (show that for such $x$, $\frac{x^{2k}}{2k} \geq \frac{x^{2k+1}}{2k+1}$, so that you are subtracting positive terms from $x$ in the computation).
But for $x\gt 1$ the approach does not work, since the Taylor series no longer converges. If you want to show that $\log(1+x)\leq x$, then you can instead show that $1+x \leq e^x$, by taking exponentials. Again you can use the Taylor series, this time for $e^x$: $e^{x} = 1 + x + \frac{x^2}{2!} + \cdots + \frac{x^n}{n!} + \cdots$ which makes $1+x\leq e^x$ for $x\geq 0$ immediate. This would establish the second of your inequalities as written at the current time, by squaring both sides of $\log(1+x)\leq x$ (since $\log(1+x)\geq 0$ when $x\geq 0$).
For $(\log(1+x))^2\leq x$, for $0\leq x\lt 1$ this follows from the first inequality by squaring, since $x^2\leq x$ on that region. For $x\gt 1$, a different approach would be required.