Showing $f:\mathbb{R^2} \to \mathbb{R}$, $f(x, y) = x$ is continuous Let $(x_n)$ be a sequence in $\mathbb{R^2}$ and $c \in \mathbb{R^2}$.
To show $f$ is continuous we want to show if $(x_n) \to c$, $f(x) \to f(c)$.
As $(x_n) \to c$ we can take $B_\epsilon(c)$, $\epsilon > 0$ such that when $n \geq$ some $N$, $x_n \in B_\epsilon(c)$.
As $x_n \in B_\epsilon(c)$ this implies that $f(x_n) \in f(B_\epsilon(c))$.
This holds for all $\epsilon$, so as $\epsilon \to 0$ and $B_\epsilon(c)$ becomes infinitely small, we can always find $n \geq$ some $N$ such that $x_n \in B_\epsilon(c)$ and $f(x_n) \in f(B_\epsilon(c))$.
Hence as $\epsilon \to 0$, $(x_n)$ clearly converges to $c$ and $f(x_n)$ clearly converges to $f(c)$.
Does that look ok?
 A: Well, since you don't use the definition of $f$ anywhere, you could substitute any function, and this would prove that it's continuous. What you need to show is that for any $\varepsilon>0$ there exists an $N$ such that $f(x_n)\in B_\varepsilon (f(c))$, instead of $f(x_n)\in f(B_\varepsilon (c))$, for every $n\ge N$.
A: Purely topological proof:
This proof works for any space $X$. In your case $X=\mathbb{R}$. You want to prove that the first projection $f:X \times X \to X$ is continuous. A function $f$ is continous if the inverse image of any open is open. So let $U \subseteq$ X be any open. Then the inverse image $f^{-1}(U)=U \times X$, which is open by definition of the product topology (its base consists of $V \times W$ with $V,W$ open in $X$). 
A: There's a bit of repetition when you say $x_n \in B_\epsilon(c) \implies f(x_n) \in f(B_\epsilon(c))$. While this is true as you define it, repeating it doesn't add to the proof. What you need to show is that the image of $B_\epsilon(c)$ is itself an open neighborhood of $f(c)$.
Another look, which uses uniform continuity...
For any open neighborhood $N_\delta(\mathbf{x})$ of $\mathbf{x} = (x_1,y_1) \in \Bbb R^2$ and $(x_2,y_2) = y \in N_\delta(\mathbf{x})$ with the usual metric we have
$$\delta > d(\mathbf{x},\mathbf{y}) = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2} \ge \sqrt{(x_1-x_2)^2} = |x_1-x_2| = d(f(\mathbf{x}),f(\mathbf{y})).$$
Thus, we can set $\delta = \varepsilon$ and obtain uniform continuity on any open subset of $\Bbb R^2$.
