Q:

Am I taking the "true meaning" of "$\subset$" and in the context of this question to be wrong? Or the question is incorrect itself?

I came across this question in my assignment.
Question 1. Let $A=\{\emptyset,\{\emptyset\},1,\{1,\emptyset\},2\}$. Does the following hold true for the above set?
$B=\{\emptyset,\{\emptyset\},\{1,\emptyset \} \} \subset A$ ? (It is told to be true in the soultions but I think it is not.)
I thought about it. Compared it with the set A but that took me nowhere. What I think is that it is false because...
1. In order to be a subset of a set (A here), the "original elements" are taken out from the "original set" (again A) and are put into these $\{$  $\}$ which converts a respective element into a subset of the set.
Now if I compare  set A and  the other set B (which I have to tell about if it holds true or not with respect to set A), I find...
(A).   $\emptyset$ is a subset of every set so B should have $\emptyset$ contained in it. (Yes!)
(B). $\emptyset$ is taken out from set A and has been placed in set B so it should be in brackets. (yes it is!) like this $\{\emptyset\}$ 
(C). Coming to third and the last element of set B which is $\{1,\emptyset\}$, I find 1 when taken out from set A should simply have been placed like this $\{1\}$ but it is not rather it has been placed like this $\{1,\emptyset\}$ which I think is incorrect. (But soluions say it is correct, which I don't agree to).
(D). If they were have to be placed in B then I think there could have been a better way which is this...  $B=\{\emptyset,\{\emptyset\},\{1\},\{\{\emptyset \}\} \} \subset A$ (True?) Or this $A=\{\emptyset,\{\emptyset\},1,\{\{1,\emptyset\}\},2\}$ (True?)
Am I correct in my approach? Also If you would like to suggest me something, go ahead.

A:

The set $B$ has $3$ elements, namely
$$\emptyset,\;\;\{\emptyset\},\;\;\{1,\emptyset \}$$ 
The set $A$ has $5$ elements, namely
$$\emptyset,\;\;\{\emptyset\},\;\;1,\;\;\{1,\emptyset\},\;\;2$$
Since each element of $B$ is also an element of $A$, we have $B \subseteq A$.

But $B \ne A$, hence $B \subset A$.

