Number of path components for products; Is my conjecture right? For two topological spaces, I am wondering how the product of the two would make the number of path components in them...

Let $X,Y$ be topological spaces and say there are $n$ and $m$ path components of them respectively(say we deal with finite $n,m$ here). What is the path components of the product $X \times Y$?

I think, it's simply $nm$.
This is because, $X$ has path $\alpha_{xy}$ for some $x,y \in X$ and $Y$ has $\beta_{ab}$ for $a,b \in Y$. Say the same paths apply to path components $x,y$ i.e. all other elements of $X$ that are path related to $x$ and the same for elements of $Y$. 
Then, well $(x,a) \in X \times Y$ for each $x \in X,a \in Y$.Then, we can define a path $I \to X \times Y$ by $f:I \to X \times Y; t \to (\alpha(t),\beta(t))$. This $f$ is continuous since $\alpha, \beta$ are paths and thus continuous(proof omitted to keep the focus on track for this problem).
So, when we think about path components in $X \times Y$, it seems like I can think $x,a$ in turns, so, for each $x \in X$ that are in a certain path component of $X$, I'll have some $\alpha$ fixed while I consider $a \in Y$ in different path components in $Y$. There are $m$ such possibilities, and now, for $x \in X$, there are $n$.(I sound rather tedious and verbose don't I...)
So for $(x,a) \in X \times Y$, there are $nm$ possibilities of path components...
This isn't a rigorous poof but is my thoughts okay? Or completely wrong?

Yes, you've got just the right idea. Let's denote the path component of $x$ in $X$ by $[x]$.
Then if we pick a point $(x,y) \in X \times Y$, and $(x', y') \in [(x,y)]$ then we have a path $\gamma: I \to X \times Y$ from $(x, y)$ to $(x', y')$. Projecting $\gamma$ onto $X$ and $Y$ gives you paths from $x$ to $x'$ and $y$ to $y'$ respectively.
Conversely, if $\alpha: I \to X$ is a path from $x$ to $x'$ and $\beta: I \to Y$ is a path from $y$ to $y'$ then $\gamma: I \to X \times Y; t \mapsto (\alpha(t), \beta(t))$ is a path from $(x, y)$ to $(x', y')$.
Therefore $[(x,y)] = [(x', y')]$ if and only if $[x] = [x']$ and $[y] = [y']$. In other words there's a bijection $\pi_0(X \times Y) \cong \pi_0(X) \times \pi_0(Y)$, where the set of all path components of $X$ is denoted by $\pi_0(X)$.
