Let $A$ be the set $\{\propto,\{1,\propto\},\{3\},\{\{1,3\}\},3\}$.
The statement "$\{1,\propto\}\subseteq A$" is false.
(Taken from A Concise Introduction to Pure Mathematics)
But I think this is true as can see $\{1,\propto\}$ as being a subset of $A$ ?
No. $\{1, \propto\}$ is an element of $A$, but not a subset of $A$. If it was a subset, every element of it, i.e. both $1$ and $\propto$, would have to be elements of $A$.
$\propto$ is an element of a, but $1$ is not. It is only contained in an element (this element is $\{1,\propto\}$ and is, by chance, also a set) so
$$1 \notin A \Rightarrow \{1,\propto\} \not\subset A$$
On the other hand, $\{3\}$ is both an element of $A$ (the third one in your list) and a subset of $A$, because all elements of $\{3\}$ (the only one being $3$) are also an element of $A$ ($3$ is the fifth in your list)
To help clarify, here are the elements of $A$:
$$\propto\\ \{1,\propto\}\\ \{3\}\\\{\{1,3\}\}\\ 3$$
The subsets of $A$ are sets consisting of elements of $A$, i.e.
$$\emptyset\\ \{\propto\}\\ \{\{1,\propto\}\}\\ \{\{3\}\}\\ \{\{\{1,3\}\}\}\\ \{3\}\\ \{\propto, \{1,\propto\}\}\\ \ldots\\ \{\{3\},3\}\\ \ldots\\ \{\propto, \{1,\propto\}, \{3\}, \{\{1,3\}\}, 3\} = A$$
"element" and "subset"; see $x\in \{\{\{x\}\}\}$ or not?
$\{1,\propto\}\subseteq A$ is false.
$\{\{1,\propto\}\}\subseteq A$ is true.
$\{1,\propto\}\in A$ is true.
No. $\{1,\alpha\}$ is a subset of $A$, if all its elements (namely $1, \alpha$) are elements of $A$, so we have to check whether $1 \in A$ and $\alpha \in A$. The elements of $A$ are
(1) $\alpha$, (2) $\{1,\alpha\}$, (3) $\{3\}$ and (4) $\bigl\{\{1,3\}, 3\bigr\}$
So, $\alpha \in A$ ($\alpha$ is No. 1 in our list above, but $1 \not\in A$. So, as not all elements of $\{1,\alpha\}$ are elements of $A$, we have $\{1,\alpha\} \not\subseteq A$.
Note that $\{1,\alpha\}$ is an elements of $A$ (no. 2 above), but this has nothing to do with it being a subset.
No, $\{1,\propto\}$ is not a subset of $A$.
In order for $S$ to be a subset of $A$, it must be true that every element of $S$ is also an element of $A$. In your case, $1$ is an element of $\{1,\propto\}$, but $1$ is not an element of $A$, so $\{1,\propto\}$ is not a subset of $A$.
If it was a subset, then $1$ would had to be an element of $A$, but it isn't. There's only a set which contains $1$, which is an element of $A$, but not $1$ itself. However, {1,∝} is an element of $A$.
