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?

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 $\Bbb R^2$.
