Please let me refer you to:
Example 4.18. The skew line $f: \mathbb R \to S^1 \times S^1$ $ f(t) = (e^{it}, e^{i\alpha t}). $ If $\alpha$ is irrational then the image of $f$ is dense in $S^1 \times S^1$, so if $V$ is an open neighborhood of $f(t)$ in $S^1 \times S^1$, then $ \overline{V \cap f(\mathbb R)} = V $ so $V \cap f(\mathbb R) \neq f(U)$.
(Source: http://www.math.toronto.edu/mat1300/smoothmaps.4.pdf, Page $16$, example $4.18$.)
Why is the image of the skew line dense (assuming $\alpha$ is not rational)?