Is $\{\varnothing\}\subset\{1,2,3\}$ true? $\{\varnothing\}\subset\{1,2,3\}$ I don't know if this is true or not. I know the emptyset is a subset of every set but this one is really confusing me.
 A: There's an analogy that might help here.  It's a little dangerous, because it's tempting to extend it where it doesn't belong, but for something like this, it might help.
Think of sets as boxes.  The things in a box are that box's members or elements.  In contrast, one box is a subset (a sub-box, if you will) of another box if and only if the second box has everything the first box has.  The empty set is an empty box, of course.
So, on the left (say), you have a box, whose only contents are an empty box.  On the right, you have another box, whose only contents are the numbers $1$, $2$, and $3$.  The box on the left is not a subset of the box on the right, because the box on the left has an empty box, which is not one of the things in the box on the right.
That empty box is a subset of the box on the right, because there's nothing in it that isn't also contained in the box on the right.  (That is, of course, because there's nothing in the empty box at all!)
Analogously, the empty set is a subset of the set $\{1, 2, 3\}$ (and is, in fact, a subset of any set at all), but the set containing the empty set is not a subset of the set $\{1, 2, 3\}$.  It is not a subset of any set that doesn't explicitly contain the empty set as an element.

Of course, it's entirely possible that this metaphor is confusing and/or annoying to you.  It happens.  But it helps some people (including me, sometimes).

ETA: It occurs to me, after I re-read your question, that you chose a curious example that just happens to present a cute little coincidence.  In Zermelo-Frankel (ZF) set theory, it is standard to define the natural numbers $\mathbb{N} = \{1, 2, 3, \ldots\}$ as sets.  In particular, we define zero as the empty set, and then every succeeding number is the union of the previous number and the set containing the previous number.  (There are technical reasons for doing it this way that needn't concern us here.)
So, therefore, we have
$$
0 = \emptyset
$$
$$
1 = 0 \cup \{0\} = \emptyset \cup \{\emptyset\} = \{\emptyset\}
$$
$$
2 = 1 \cup \{1\} = \{\emptyset\} \cup \{\{\emptyset\}\}
                 = \{\emptyset, \{\emptyset\}\}
$$
$$
3 = 2 \cup \{2\} = \{\emptyset, \{\emptyset\}\}
                   \cup \{\{\emptyset, \{\emptyset\}\}\}
                 = \{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}
$$
So the assertion you originally wrote was
$$
\{\emptyset\} \subset \{\{\emptyset\}, \{\emptyset, \{\emptyset\}\},
                      \{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}\}
$$
which turns out to be false.  But if you had made just a small change and asked if $\{\emptyset\} \subset \{0, 1, 2, 3\}$, that would have been true (!).  It turns out that if you define the natural numbers this way, then one number is less than another (in the informal sense) if it is a proper subset of the other (in the formal sense).  So "of course" the changed assertion is true, because $1 < 4$.  But $\{1, 2, 3\}$ isn't a number in that definitional framework.
But don't reason this way ordinarily.  This was just something that you happened to pick that fit.  People don't ordinarily work with numbers as sets; it's useful to know that it can be done that way, rigorously, but that's not how arithmetic is performed.
A: $\{\emptyset\}$ is not a subset of $\{1,2,3\}$ because $\emptyset$ is not an element of $\{1,2,3\}$.
But it is a subset of any set that does have $\emptyset$ as an element, such as $\{\emptyset , bananas \}$
