$\mathbb{R}^n$ as the product of $n$ copies of the real line: the product topology is the Euclidean topology and the projections are continuous, open and surjective

Fix $n \in \mathbb{N}$ with $n \ge 1$ and give $\mathbb{R}$ its usual topology (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y) = |x-y|$ is a metric, its open balls are the intervals $(x-r, x+r)$, and it is unbounded, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not). Let $\mathbb{R}^n = \prod_{k<n} \mathbb{R}$ carry the product topology (The product set $\prod_{i \in I} X_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space), with projections $\pi_j(x) = x_j$. Then:

1. The product topology is the Euclidean topology. It is the metric topology of $d_\infty$, and equally of $d_1$ and of $d_2$ ($\mathbb{R}^n$ as the set of functions $n \to \mathbb{R}$, and $d_1$, $d_2$, $d_\infty$ are metrics on it, For $n \ge 1$ the product topology on $n$ copies of the usual topology of $\mathbb{R}$ is the metric topology of $d_\infty$ on $\mathbb{R}^n$, and hence also of $d_1$ and $d_2$, so $\mathbb{R}^n$ as a product and $\mathbb{R}^n$ as a metric space are one space); in particular $\mathbb{R}^n$ with the product topology is metrizable, and "open in $\mathbb{R}^n$" has one meaning.
2. A basis of open boxes. The sets $\prod_{k<n}(a_k, b_k)$ with $a_k < b_k$ for every $k < n$ form a basis (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length), since the $d_\infty$-ball $B(x,r)$ is exactly the box $\prod_{k<n}(x_k - r,\ x_k + r)$.
3. The projections are continuous, open and surjective. Continuity and openness are the general facts (A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice, claims 1 and 3; Continuity of a map of topological spaces at a point and globally, Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological). Surjectivity needs no choice principle here: for $t \in \mathbb{R}$ the constant function $x$ with $x_k = t$ for every $k<n$ satisfies $\pi_j(x) = t$.
4. Componentwise continuity. For a space $Z$, a function $h : Z \to \mathbb{R}^n$ is continuous if and only if each of its $n$ components $h_k = \pi_k \circ h : Z \to \mathbb{R}$ is (A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice, claim 2). This is the statement usually quoted as "a vector-valued map is continuous exactly when its coordinate functions are", and here it is a special case of a theorem about arbitrary products.

• This is the example the seam lemma exists for. Without For $n \ge 1$ the product topology on $n$ copies of the usual topology of $\mathbb{R}$ is the metric topology of $d_\infty$ on $\mathbb{R}^n$, and hence also of $d_1$ and $d_2$, so $\mathbb{R}^n$ as a product and $\mathbb{R}^n$ as a metric space are one space the symbol $\mathbb{R}^2$ would name two spaces on these two pages, the product of two lines and the metric space of $\mathbb{R}^n$ as the set of functions $n \to \mathbb{R}$, and $d_1$, $d_2$, $d_\infty$ are metrics on it, and every sentence about open subsets of it would be ambiguous. Claim 1 says the two are one space, so the hyperbola, the square and the Sorgenfrey plane below may each be discussed in whichever language is shorter.

• Surjectivity is free here and is not free in general. For an infinite index set, surjectivity of a projection is the Axiom of Choice (A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice, claim 4). For $\mathbb{R}^n$ the constant function does the work, and the same trick works for any product of copies of one nonempty space, over any index set.

• The Euclidean metric plays no role in the topology. All three of $d_1$, $d_2$ and $d_\infty$ induce the product topology, so nothing topological about $\mathbb{R}^n$ singles out $d_2$; what singles it out is metric structure, such as which sets are balls, and that is not visible to the topology (Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not).