How can a subspace have a lower dimension than its parent space? 
If $V$ is a vector subspace of $W$, then
$$\dim(V) \le \dim(W)$$

Why? Does that mean that for
$$W = \mathbb{R}^3\\
V = \{(0,0)\}$$
$V$ is a valid subspace of $W$? But $V$ only has two coordinates, and $W$ has $3$...
I've always been under the impression that a subspace's dimension should be equal to its parent space's dimension.

EDIT
I made a mistake with my example - the answer is no (as you all pointed out), however, the real core question is, how can a subspace have a lower dimension than its parent space?
 A: No. $\{(0,0)\}$ is not a subspace of $\mathbb{R}^3$. 
A subspace must be a subset of its parent vector space. 
Now, $\{(0,0,0)\}$ is a subspace of $\mathbb{R}^3$ and $\{(0,0)\}$ is a subspace of $\mathbb{R}^2$, but $\{(0,0)\}$ is not a subspace of $\mathbb{R}^3$.
As for your other question, a subspace's dimension cannot exceed its parent's dimension, but it by no means must be equal to it.
For example: The subspaces of $\mathbb{R}^3$ are...
$\mathbb{R}^3$ itself (every vector space is a subspace of itself).
Any plane through the origin is a 2-dimensional subspace of $\mathbb{R}^3$.
Any line through the origin is a 1-dimensional subspace of $\mathbb{R}^3$.
The origin itself, $\{(0,0,0)\}$ is the 0-dimensional subspace of $\mathbb{R}^3$.
Why must the dimension of a subspace not exceed the dimension of it's parent? It's essentially because any linearly independent subset (like a basis for a subspace) can be extended to a basis for the whole vector space.
Edit: Coordinates written down to represent a subspace $\not=$ the dimension of the subspace.
Example: $W = \{(c,2c,3c)\;|\; c \in \mathbb{R}\}$ is a subspace of $\mathbb{R}^3$. Now, yes, elements of $W$ are 3-tuples, but this does not make $W$ itself 3-dimensional. 
Think of "dimension" as meaning the minimum number of parameters needed to describe the subspace, so for $W$ this is "1". Notice that $W$ consists of multiples of $(1,2,3)$. This means that $\{(1,2,3)\}$ is a basis for $W$ and thus (since the basis has only 1 element), $W$ is a 1-dimensional subspace of $\mathbb{R}^3$.
Now what if all you know is the world of $W$? Then you really don't need a 3-tuple to know who you are. Instead of saying, "Hey I'm $(2,4,6)$." You could just say, "Hey I'm 2." (as long as everyone knows that's shorthand for $2(1,2,3)=(2,4,6)$. 
Another analogy is like driving on a highway. If someone asks you where you are on I-40 in North Carolina, you don't need to give them your latitude and longitude. You could just say, I'm at mile marker 319. Even though the highway is a 3-dimensional thing (going different directions at different altitudes) in some sense (internally) it's really 1-dimensional. That internal kind of dimension is the one being discussed here.
I hope this helps!
A: Sub- just means within. 
-space means when viewed in isolation from the parent space, it is a vector space in its own right. 
In using the term "subspace", there is no implication that the subspace has to have the same dimension as the parent space.

Also, you are confusing what dimension means. The set $\{(0,0,0)\}$ has dimension zero, because it is just a point. The fact that we are discussing an element that is expressed as an ordered triple is irrelevant to discussing dimension. At best, that just tells us our space is a subspace of $\mathbb{R}^3$.
A: I've found it useful not to picture a space in its totality. Try to understand what a space is by its bases or spanning sets. Suppose $V$ is a vector space with dimension $n$. That is, any $n$ linearly independent vectors in $V$ will span it. Now consider a linearly independent set of vectors in $V$ with less than $n$ vectors. Clearly the subspace spanned by this set has a smaller dimension, by definition. 
Now for your question. Consider any set of vectors in $V$ with more than $n$ vectors. You might have learnt that this set cannot be linearly independent and cannot be a basis for $V$. Furthermore every subspace in $V$ will have a spanning set and hence a basis. This discussion should tell you that the dimension of such subspaces must be less than that of the whole space since a basis for such a set will contain $n$ or less vectors. 
