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Show that the subspace $(a, b)$ of $\mathbb{R}$ is homeomorphic with $(0, 1)$

In this question (taken form Munkres), is $(0, 1)$ referring to the following subspace $\left((0, 1) \subset \mathbb{R}\ , \ \mathcal{T} = \{(0, 1) \cap U \ | \ U \in \mathcal{T}_{\mathbb{R}}\}\right)$?

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    I'd suggest proving that $x \to x-a$ is a homeomorphism of $\mathbb{R}$ to itself. Similarly I would then show that $x \to \frac{x}{b-a}$ is a homeomorphism of $\mathbb{R}$ to itself. The restriction of the composition of these maps is a homeomorphism from $(a,b)$ to $(0,1)$. – 2017-01-16

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Yes, $(0,1)$ refers to the subset $(0,1)\subset\mathbb{R}$ with the subspace topology. In general, subsets of "standard" spaces (like $\mathbb{R}$ or $\mathbb{R}^n$) are commonly assumed to have the subspace topology unless stated otherwise.