question regarding proof for the dimension of the sum of subspaces theorem: "Given finite dimensional subspaces $U_1$ and $U_2$ in a vector space V then: $$dim(U_1+U_2)=dim(U_1)+dim(U_2)-dim(U_1 \cap U_2)$$ "
The first step of the proof establishes that $U_1 \cap U_2$ is finite dimensional. 
My question is why exactly is this important? Is it because the dimension of a non-finite dimensional space is undefined??
(I gather that this is of course important, just want to know the reason)
 A: Dimension is meaningful even when it is not finite. In general, the dimension of a vector space is defined to be the cardinality of its basis (one can show that any two bases have to share the same cardinality). 
Cardinality essentially means the "size" of a set. For example: the cardinality of $\{a,b,c\}$ is 3, whereas the cardinality of $\mathbb{N} = \{ 0,1,2,\dots \}$ is $\aleph_0$ (i.e., countable infinity). The cardinality of $\mathbb{R}$ is $2^\aleph_0$ (i.e., continuum). We have (this can be made precise) that $3<\aleph_0<2^\aleph_0$. So that each of my example sets is strictly bigger than the previous one.
If you take polynomials $\mathbb{R}[x] = \{ n \geq 0 \text{ and } a_nx^n +\cdots+a_1x+a_0 \;|\; a_0,\dots,a_n \in\mathbb{R}\}$, you get an infinite dimensional vector space (over the reals). So many texts write:  $\dim(\mathbb{R}[x])=\infty$. 
If we are being more careful, we will specify the kind of infinite dimension we have. Specifically, $\dim(\mathbb{R}[x])=\aleph_0$. The reason for this is that $\beta = \{1,x,x^2,\dots\}$ forms a basis for the space of polynomials and there is an invertible map between $\beta$ and $\mathbb{N}$ (i.e., $x^i \mapsto i$). 
Now while subtraction is ok for finite quantities, it is no longer ok for cardinalities. This is much like the fact that $\infty-\infty$ is an indeterminant form. 
For example: $\{1,2,3,\dots\}$ take away $\{3,4,\dots\}$ leaves $\{1,2\}$, so one might conclude that $\infty-\infty=2$. But if we took $\{1,2,3,\dots\}$ and removed $\{2,4,6,\dots\}$ we would have $\{1,3,5,\dots\}$ so now we should have $\infty-\infty=\infty$. 
So subtracting cardnalities is problematic. Adding them is not. One says that if $A$ has cardinal number $\alpha$ and $B$ has cardinal number $\beta$ then the disjoint union of $A$ and $B$ (put these sets together without allowing overlapping) has cardinal number $\alpha+\beta$.
For example: $A=\{1,2,3\}$ and $B=\{1,2\}$ then make $B$ distinct: $B' = \{1',2'\}$ put them together and get $\{1,2,3,1',2'\}$ thus $2+3=5$. 
One can make sense of this for infinite stuff as well.
So why did you text prove finite dimensional first? Possibly because the statement involved a subtraction. 
Actually if you say: $\dim(U_1+U_2)+\dim(U_1\cap U_2) = \dim(U_1)+\dim(U_2)$, then this statement holds without the finite dimensional assumption!
Proof sketch: Let $\alpha$ be a basis for $U_1 \cap U_2$. Extend this to a basis $\beta = \alpha \cup \beta_0$ (disjoint union) for $U_1$ and extend also to a basis $\gamma = \alpha \cup \gamma_0$ (disjoint union) for $U_2$. 
You can show that $\alpha \cup \beta_0 \cup \gamma_0$ (disjoint union) is a basis for $U_1+U_2$. So that:
$$\dim(U_1+U_2)+\dim(U_1\cap U_2)=|\alpha \cup \beta_0 \cup \gamma_0| + |\alpha|$$ $$= |\alpha|+|\beta_0|+|\gamma_0|+|\alpha| = |\alpha \cup \beta_0| + |\alpha\cup\gamma_0| = \dim(U_1)+\dim(U_2)$$
[where $|X|$ denotes the cardinality of $X$. We used the fact that $|X\cup Y| = |X|+|Y|$ for disjoint sets and that cardinal addition is commutative and associative.]
This proof works whether the dimensions are finite or infinite. :)
