Let us first show that $\ln 2 \lt 1$. We have $\ln 2=\int_1^2\frac{dt}{t}$.
Divide the interval from $1$ to $2$ into $1$ part. The upper Riemann sum is the width of the interval, times the value of the function $\frac{1}{t}$ at $t=1$. This upper Riemann sum is $1$, and is clearly bigger than the integral.
Now let us show that $\ln 4 \gt 1$. We have $\ln 4=\int_1^4\frac{dt}{t}$.
Divide the interval from $1$ to $4$ into $3$ equal parts, and find the corresponding lower Riemann sum. The minimum of $\frac{1}{t}$ on the first part is $\frac{1}{2}$. For the other two parts, the minima are $\frac{1}{3}$ and $\frac{1}{4}$. So the lower Riemann sum is $\ge \frac{1}{2}+\frac{1}{3}+\frac{1}{4}$. This is already $\gt 1$, so the integral is $\gt 1$.
We could alternately note that $\ln 4=2\ln 2$. Then we can divide the interval from $1$ to $2$ into $1$ part, and note that our function value is $\frac{1}{2}$ at the right endpoint, and $\frac{1}{t}$ is decreasing. So the lower Riemann sum is $\frac{1}{2}$, and therefore $\ln 2 \gt \frac{1}{2}$, so $2\ln 2 \gt 1$. But this is not quite in the spirit of the game, since we have used a property of $\ln$, namely $\ln 4=2\ln 2$. This would first have to be established from the definition of $\ln$ as an integral.
Remark: We were a little casual about concluding that the integral from $1$ to $2$ is less than $1$. In principle we only showed that it is $\le 1$. If you are very fussy, you can divide the interval $[1,2]$ into two equal parts. Then the upper sum is $\frac{1}{2}\left(\frac{1}{1}+\frac{1}{3/2}\right)\lt 1$.
The above calculations will only make full sense if one draws a picture of $y=\frac{1}{t}$, and visually identifies the upper and lower sums mentioned. They are all either the area of a rectangle, or the sum of the areas of a small number of rectangles.