Why is $\{\{1\}\}$ not equal to $\{1,\{1\}\}$?

Think of $A$ as a bag which contains within it another smaller bag with a one in it.

$A=\underbrace{\{~~~~~~~\overbrace{\{1\}}^{\text{second bag}}~~~~~~~~\}}_{\text{first bag}}$

On the other hand, $B$ is a bag which contains in it not only a second bag with a one in it, but also a one which is loose.

$B=\underbrace{\{~~~~~~~~\overbrace{\{1\}}^{\text{second bag}}~~~~~\overbrace{1}^{\text{this too}}~~~~~~~\}}_{\text{first bag}}$

$1\in B$ but $1\not\in A$. There is no "loose 1" in $A$, there is only a bag with a one in it in $A$.

Thus, $A\neq B$

You're correct that all elements in $A$ are in $B$, but not the other way around - $B$ includes the element $1$, but $A$ only has $\{1\}$. Think of it like boxes - $B$ is a box that includes one item and also a box that itself contains one item; $A$ is just a box containing a box containing an item.