Question: The following statement appears as a true or false question:

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> $\{ \varnothing \} \subseteq A$ for all sets $A$.
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> 
> 

Because it is the set containing the null set, instead of just the null set itself, I'm hesitant to say whether this is true or false. Can anyone clarify for me?

Answer: 
You're rightfully reluctant.

We have $\varnothing\subseteq A$ for all sets $A$, but the set *containing* the null set as an element, namely, $\{\varnothing\}$ is **not** a subset of *every* set. 

For example, given $$A = \{1, 4, 9, 16\},$$ 

* we have $1\in A,\;$ so $\;\{1\} \subseteq A$,
* while we do have $\varnothing \subseteq A,\,$ $\,\varnothing \notin A\,$ so $\,\{\varnothing\} \not\subseteq A$.

So it is ***not*** the case that *for all* sets $A$, $\,\{\varnothing\} \subseteq A$.

**Remark:** We can construct a set $A$ such that $\{\varnothing\} \subseteq A$. Put $A = \{\varnothing\}$. Then $\varnothing \subseteq A$ AND $\varnothing \in A$. So it follows that $\{\varnothing\} \subseteq A$.
$\{\}$ is not the same as $\{\varnothing\}$.

$\{\}$ is the empty set; it contains no elements. Another way of writing an empty set is $\varnothing$. 

However, $\{\varnothing\}$ is not empty; it contains the empty set. Therefore $\{\varnothing\}$ is not an empty set.