Prove that a function $f: \mathbb{R}^2 \to \mathbb{R}$ is continuous Here's the full question:
Let $f:\mathbb{R}^2 \to \mathbb{R}$ be a function defined as follows:
$$\forall x \in \mathbb{R}^2: f(x) = x_1+x_2$$
where $x = (x_1,x_2)$. Now, define a metric $d$ on $\mathbb{R}$ by the following correspondence:
$$\forall x,y \in \mathbb{R}: d(x,y) := |x-y|$$
and a metric $d'$ on $\mathbb{R}^2$ by the following correspondence:
$$\forall x,y \in \mathbb{R}^2: d'(x,y) = \sqrt{(x_1-y_1)^2+(x_2-y_2)^2}$$
Prove that $f$ is continuous.

Proof Attempt:
Let $a = (a_1,a_2) \in \mathbb{R}^2$ be fixed but arbitrary. We will prove continuity at $a$. Let $\epsilon > 0$ be given. We need a $\delta > 0$ such that:
$$d'(x,a) < \delta \implies d(f(x),f(a)) < \epsilon$$
$$\iff \left( \sqrt{(x_1-a_1)^2+(x_2-a_2)^2} < \delta \implies |f(x)-f(a)| < \epsilon \right)$$
Now, consider the following list of inequalities:
$$|f(x)-f(a)| = |(x_1+x_2)-(a_1+a_2)| = |(x_1-a_1)+(x_2-a_2)|$$
$$|f(x)-f(a)| \leq |x_1-a_1| + |x_2-a_2| = |x_1-a_1| \cdot 1 + |x_2-a_2| \cdot 1$$
By the Cauchy-Schwarz Inequality, it follows that:
$$|f(x)-f(a)| \leq \sqrt{2} \cdot \sqrt{(x_1-a_1)^2+(x_2-a_2)^2} < \sqrt{2} \delta$$
Define $\delta := \frac{\epsilon}{\sqrt{2}}$. Then, the required implication follows. This proves that $f$ is continuous at $a$ and since our choice of $a$ was arbitrary, it follows that $f$ is continuous.
Does this argument work? If it doesn't, then why? How can I fix it?
 A: Your proof sounds good. Here I propose another way to prove it based on the sequential characterization of continuity. In order to do so, we shall need the following result:
Proposition 1
Let $(x_{n},y_{n})$ be a sequence in $\mathbb{R}^{2}$. Then $(x_{n},y_{n})$ converges to $(x,y)$ iff $x_{n}$ converges to $x$ and $y_{n}$ converges to $y$.
Proposition 2
Let $(X,d_{X})$, $(Y,d_{Y})$ be metric spaces and $x_{0}\in X$. Then
(a) $f:X\to Y$ is continuous at $x_{0}$ iff
(b) whenever $x_{n}\in X$ is a sequence that converges to $x_{0}$ in accordance to $d_{X}$, then $f(x_{n})$ converges to $f(x_{0})$ in accordance to $d_{Y}$.
Solution
Let $(x_{n},y_{n})\in\mathbb{R}^{2}$ be a sequence which converges to $(x,y)$. Then $x_{n}$ converges to $x$ and $y_{n}$ converges to $y$. Consequently, due to the properties of convergent sequences, one has that
\begin{align*}
f(x_{n},y_{n}) = x_{n} + y_{n} \Rightarrow \lim_{n\to\infty}f(x_{n},y_{n}) = x + y = f(x,y)
\end{align*}
whence we conclude that $f(x,y) = x + y$ is continuous.
Hopefully this helps.
