showing inclusion map from topological space to product space is continuous let $i:X \rightarrow X \times Y$ where $i(x)=(x,y_{0})$. I am to show that $i$ is continuous. ($X,Y$ topological spaces of course)
I started off by looking at an example $i:\mathbb{R} \rightarrow \mathbb{R}\times\mathbb{R}$ with $i(x)=(x,y_{0})$ where it is clear if one is to take an open ball in $\mathbb{R}\times\mathbb{R}$ and fix a $y_0$ the the ball becomes an open interval and from there it is clear the $i$ is continuous. (that is the inverse images of the open balls are clearly open intervals, since the balls are basically already intervals via $i$)
Although I am having trouble taking this concept to abstraction to prove the statement. A hint as to how to do this would be greatly appreciated.
 A: I don't believe it is helpful to restrict oneself to the example where e.g. $X=Y=\mathbb R$. The result is more fundamental than that. 
Let $U$ be open in $X\times Y$, then by definition of product topology we may write $$U = \bigcup_{\alpha\in I} (U_\alpha^X\times U_\alpha^Y)$$ where $U_\alpha^X$ is open in $X$ and $U_\alpha^Y$ is open in $Y$ for each $\alpha\in I$. If $y_0\notin U$ then $i^{-1}(U)=\varnothing$ is open in $X$. If $y_0\in U$, let $J\subset I$ be the set of $\beta$ such that $y_0\in U_\beta^Y$. Then
\begin{align}
i^{-1}(U) &= i^{-1}\left(\bigcup_{\alpha\in I} (U_\alpha^X\times U_\alpha^Y \right)\\
&= i^{-1}\left(\bigcup_{\beta\in J} (U_\beta^X\times U_\beta^Y \right)\\
&= \bigcup_{\beta\in J} i^{-1}(U_\beta^X\times U_\beta^Y)\\
&= \bigcup_{\beta\in J} U_\beta^X
\end{align}
is open in $X$ as the union of open sets in $X$. It follows that $i$ is continuous.
A: A general fact about any product topological space:

A function $Y \to \prod_{i \in I} X_i$ from any space $Y$ to a product space is continuous iff $\forall i\in I: \pi_i \circ f: Y \to X_i$ is continuous, where the $\pi_i$ are the standard projections on the product. 

This is the universal mapping property which holds for all initial topologies, of which products are an important special case.
For your $i$, we have that $\pi_1 \circ i = 1_X: X \to X$, the identity on $X$, which is always continuous, and $\pi_2 \circ i = c_{y_0}: X \to Y$ the constant map with value $y_0$, which is also continuous, as any constant map (all inverse images of (open) sets are $Y$ or $\emptyset$, always open). So $i$ is continous. That is the whole proof.
