Fix two real numbers $a$ and $b$, $0<a<b$. Define $f: \mathbb{R}^2 \rightarrow \mathbb{R}^3$ by
$$f_1(s,t) = (b+a \cos s) \cos t$$ $$f_2 (s,t) = (b+a \cos s ) \sin t$$ $$f_3(s,t) = a \sin s.$$
$f$ maps $\mathbb{R}$ onto the torus, $K = f(\mathbb{R})$.
Let $\lambda$ be an irrational. Define $g:\mathbb{R} \rightarrow K$ by $g(t) = f(t, \lambda t )$.
Prove that the range of $g$ is a dense subset of $K$.
I know that this can proven using the Kronecker approximation theorem, as shown here A mapping from $\mathbb{R}^1$ to a dense subset of the surface of torus in $\mathbb{R}^3$ .
My question is can't we simply use continuous functions to prove this. Define $h: \mathbb{R} \rightarrow \mathbb{R}^2$ by $h(t) = (t, \lambda t)$. $h$ is continuous. We also know that $f$ is continuous, so any restriction of $f$ is continuous. Then $g(t) = f(h(t))$ is continuous.
Then, by problem 4.4 in Rudin, if $g$ is a continuous mapping of a metric space $X$ into a metric space $Y$ and $E$ is a dense subset of $X$, then $g(E)$ is dense in $g(X)$. Since $\mathbb{R}$ is dense in $\mathbb{R}$, and $g$ is continuous, then $g(\mathbb{R})$ is dense in $K$.
Where is my mistake? Or is this valid?
EDIT: The mistake is in the last sentence as pointed out by Salcio in the comments.
