"Counter-example" to closed graph theorem

Question

The closed graph theorem states that the graph of a continuous function from any space $X$ to the Huasdorff space $Y$ is closed, as stated in this MSE question.

But the graph of $f: (0, \infty) \to \mathbb{R}$ given by $\sin(1/x)$ is not closed in $\mathbb{R}^2$, because it does not equal its closure.

Why is this not a counter-example?

Answer 1

The correct statement is that the graph of a continuous function $f: X \to Y$, where $Y$ is Hausdorff, is closed in $X \times Y$. The graph in your example is closed in $(0,\infty) \times \mathbb R$, though not in $\mathbb R^2$.