Prob. 10, Sec. 21, in Munkres' TOPOLOGY, 2nd ed: How to establish the continuity of this map from $\mathbb{R} \times \mathbb{R}$ into $\mathbb{R}$? The standard topology on the set $\mathbb{R}$ of all real numbers is the topology having as a basis all the open intervals $(a, b)$, where $a$ and $b$ are real numbers and $a < b$. 
So the product topology on $\mathbb{R}$ is the topology having as a basis the collection 
$$ \{ \ U \times V \ \colon \ U \mbox{ and } V \mbox{ are open in } \mathbb{R} \ \},$$
and this topology also has the following collection as a basis. 
$$ \{ \ (a, b) \times (c, d) \ \colon \ a, b, c, d \in \mathbb{R}, \ a < b, \ c < d \ \}.$$
Now let the function $f \colon \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ be defined by 
$$ f( x \times y ) \colon= x^2 + y^2 $$
for all $x \times y \in \mathbb{R} \times \mathbb{R}$. 
Do we have to use the metric space $\varepsilon$-$\delta$ argument? If so, then how to proceed if we take for $\mathbb{R}$ the metric $d(x, y) = \lvert x-y \rvert$ and for $\mathbb{R} \times \mathbb{R}$ the metric 
$$ \rho \left( \ (u, v) \ , \ (x, y) \ \right) = \max \left\{ \ \lvert u-x \rvert \ , \ 
\lvert v-y \rvert \ \right\}?$$ 
Or, can we just use the above formulations of the topologies for $\mathbb{R}$ and $\mathbb{R} \times \mathbb{R}$? And if so, then how to proceed? 
What earlier results in Munkres can we have recourse to in establishing the continuity of this map? 
 A: You can define a metric as erz suggests and use an $\varepsilon$-$\delta$ argument. This will be similar to the following argument.
To show that $f$ is continuous, we need to show that the preimage of every basic open subset of $\mathbb{R}$ is a union of the basic open subsets of $\mathbb{R}\times\mathbb{R}$.
Fix a basic open subset of $\mathbb{R}$ of the form $(a,b)$, and consider the preimage
$$
f^{-1}(a,b).
$$
We show that for each $(x_0,y_0)\in f^{-1}(a,b)$ that there is a basic open subset $V$ of $\mathbb{R}\times\mathbb{R}$ such that $(x_0,y_0)\in V\subseteq f^{-1}(a,b)$.
To this end fix $(x_0,y_0)\in f^{-1}(a,b)$.
Setting $r:=f(x_0,y_0)$, we have $r\in(a,b)$.
Denote $\varepsilon:=\min\{r-a,b-r\}$. Let
$$\delta\in\left(0,\min\left\{1,\frac{\varepsilon}{4|x_0|+2},\frac{\varepsilon}{4|y_0|+2}\right\}\right].
$$
(To see where these expressions come from, use the $\varepsilon$-$\delta$ definition of continuity to show that the function $x\mapsto x^2$ is continuous, and choose a $\delta$ which works for $\varepsilon/2$.)
Define $V:=(x_0-\delta,x_0+\delta)\times(y_0-\delta,y_0+\delta)$ and fix
$(x,y)\in V$. Then $|x-x_0|<\delta\le1$ gives
$$ |x|\le|x-x_0|+|x_0|\le1+|x_0| $$
and consequently
$$
|x^2-x_0^2|=|x-x_0||x+x_0|<\delta(|x|+|x_0|)<\delta(2|x_0|+1)
\le \frac{\varepsilon}{4|x_0|+2}(2|x_0|+1)=\varepsilon/2.
$$
Similarly $|y^2-y_0^2|<\varepsilon/2$.
Therefore
$$
|f(x,y)-r|
=|x^2+y^2-x_0^2-y^2|
\le|x^2-x_0^2|+|y^2-y_0^2|<\varepsilon,
$$
which implies by our choice of $\varepsilon$ that $f(x,y)\in (a,b)$.
Therefore $(x_0,y_0)\in V \subseteq f^{-1}(a,b)$.

Alternatively you could use sequences, which is the method I prefer.
You have to first prove the following (which I assume Munkres does):

If $X$ is first countable, then a function $f:X\to Y$ is continuous iff whenever a sequence $(x_n)$ in $X$ converges to some point $x\in X$, we have $f(x_n)\to f(x)$.

The forward direction is straight-forward (and true in general). For the reverse direction, suppose by contraposition that $f$ is not continuous at some $x\in X$. Then there is $U$ open in $Y$ such that $f(x)\in U$ and for every open neighborhood $V$ of $x$, we have $f(V)\not\subseteq U$. Letting $(V_n)$ be a decreasing neighborhood base at $x$, for each $n$ we can pick $x_n\in V_n$ such that $f(x_n)\not\in U$. Then $x_n\to x$ and $f(x_n)\not\to f(x)$.

A sequence $(x_n)$ in a product $X=\prod_{i\in I} X_i$ converges to a point $x$ iff $x_n(i)\to x(i)$ for every $i$.

This is a nice exercise for using the product topology, so I won't spoil it.
Once we have these preliminaries out of the way, suppose $((x_n,y_n))_n$ is a sequence in $\mathbb{R}\times\mathbb{R}$ which converges to $(x,y)$. Then in particular $x_n\to x$ and $y_n\to y$. Hence $x_n^2\to x^2$ and $y_n^2\to x^2$, and consequently
$$ f(x_n,y_n)=x_n^2+y_n^2\to x^2+y^2=f(x,y). $$

It's a little dishonest to claim that these proofs are really different. Indeed, the sequences proof relies on the continuity of the map $x\mapsto x^2$ and the map $(x,y)\mapsto x+y$, and the claim that a sequence converges in a product iff it converges on each component.
The latter has a analogue for continuous functions:

A function $f:X \to \prod_{i\in I}Y_i$ is continuous iff $\pi_i\circ f$ is continuous for each projection $\pi_{i_0}:\prod_{i\in I}Y_i\to Y_{i_0}$.

One can now give a proof that avoids sequences and avoids $\varepsilon$-$\delta$ estimations by writing $f=a\circ g$, where $a:(x,y)\mapsto x+y$ and $g:(x,y)\mapsto(x^2,y^2)$.
One has to verify that $a$ is continuous.
Now $g$ is continuous because $(\pi_1\circ g)(x,y)=x^2=(s\circ \pi_1)(x,y)$ is continuous, where $s:x\mapsto x^2$, and similarly $\pi_2\circ g$ is continuous.
Then we have $f$ as a composition of continuous functions.
