How are a set of matrices a topological space? For instance, is the topology on the set of all $2 \times 2$ real matrices basically $\mathbb{R}^4$
 A: There are many ways to define the topology on the space of matrices. 
For people studying riemannian geometry or lie groups, they usually identify $M_n(\mathbb{R})$ with $\mathbb{R}^{n\times n}$, so that they can use the Euclidean metric to introduce the metric tensor.
For operator theorists, the topology would follow from the operator norm
\begin{equation}\|A\|=sup\{\|Ax\|: x\in \mathbb{R}^n, \|x\|=1\},\end{equation} because this can be easily generalized to infinite dimensional spaces.
I also vaguely remember people doing numerical analysis have their own way to define a topology that help them measure error bounds in computations.
But as Nobert pointed out, on a finite dimensional vector space, there is only one natural topology in the sense that addition and scalar multiplication are continuous, so all these approaches give the essentially unique topology. This illustrates that topology is too 'sparse' sometimes and we need more rigid structures to control things due to different applications.
A: You can take any sensible topology on $\mathrm{Mat}_{m,n}(\mathbb{R})$ and you will get the usual topology of $\mathbb{R}^{mn}$. By sensible I mean topology which makes addition of matrices and multiplication by scalars continuous (this topologies are called vector topologies). 
Now we recall the following fact (see Rudin's Functional Analysis, theorems 1.19-1.21). On the $k$ dimensional vector space over $\mathbb{R}$ there is only one vector topology - topology of $\mathbb{R}^k$.
And the result follows.
A: Yes, exactly.
Topology can also be caught by limit of sequences*: a set is closed iff it is closed under limits. And the limit for $n\times m$ matrices is taken 'coordinatewise' just as in $\Bbb R^{nm}$
*at least, in separable metric spaces: those with a countable dense set (now $\Bbb Q^{n\times m}$ is such)
