The topologist's sine curve is connected

Given: The graph $C:=\{(x,\sin (1\mathrm{/}x)):0<x\le 1\}$ and the set $S:=C\cup (\{0\}\times [-1,1])$ in ${\mathbb{R}}^2$.

Every interval in the real line, including $(0,1]$, is connected (The connected subspaces of $\mathbb{R}$ with its usual topology are exactly the order-convex subsets, the published characterisation transported by the identification of the two descriptions of "open in $\mathbb{R}$").

A continuous image of a connected subset is connected (A continuous image of a connected space is connected, and connectedness is a topological property).

A map into a product is continuous exactly when each component is continuous (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).

Sine has range $[-1,1]$ (Signs, monotonicity intervals, and ranges of sine and cosine).

Sine has period $2\pi$ (The zero sets of sine and cosine and the least positive common period 2 pi).

The positive naturals are cofinal, and for every real $\epsilon >0$ some positive integer $N$ satisfies $1\mathrm{/}N<\epsilon$ (Every complete ordered field is Archimedean, For every $\epsilon >0$ in a complete ordered field there is a natural $n\ge 1$ with $1\mathrm{/}n<\epsilon$).

The reciprocal is continuous away from zero, sine is continuous, and composites of continuous real functions are continuous (Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function, The derivatives of sine and cosine are cosine and minus sine, A function differentiable at $c$ is continuous at $c$, A composite of continuous functions is continuous, with no side hypothesis of the kind the composition of limits needs).

If $A$ is connected and $A\subseteq B\subseteq \bar{A}$, then $B$ is connected (If $A$ is connected and $A\subseteq B\subseteq \bar{A}$ then $B$ is connected; in particular the closure of a connected set is connected).

The number $\pi =2\gamma$ is positive because the smallest positive zero of cosine satisfies $\gamma \in (0,2)$ (Pi as twice the smallest positive zero of cosine, Cosine has a smallest positive zero, lying strictly between zero and two).