What does it mean to equip a subset with the subspace topology? This goes for equipping any set with any topology. Suppose that I want to equip a subset $A \subseteq X$ where $X$ is a topological space, with the subspace topology. I know that by definition of the subspace topology that $\tau_A = \{U \cap A | U \in \tau\}$. But what does this do to $A$?
 A: A topology $\tau$ on a set $X$ is merely a collection of subsets of $X$ that conforms to certain rules (rules which have been chosen for their utility with respect to proving results elsewhere in mathematics$^\dagger$).
Suppose we have a set $X$ and a topology $\tau \subset \mathcal{P}(X)$. Then, for any subset $Y \subset X$, there always exists, whether explicitly acknowledged or not, what is called the subspace topology, $\tau_{_Y}$, on that subset, which is always defined with respect to an original topology, and whose open sets are, namely, where $Y$ intersects with each of the original open sets. This is to say: $U \cap Y$ is an element of $\tau_{_Y}$ whenever $U \subset X$ is an element of $\tau$.

$\underline{\textbf{Exercise}}$: Verify that $\tau_{_Y}$ as we have defined it is indeed a legitimate topology on $Y$, i.e. that it conforms to the rules a collection of subsets must follow to be a topology.

But the existence of $\tau_{_Y}$ doesn't "verb" anything to $Y$, which still exists independently as itself.

$^\dagger$If we know a given object is a particular kind of topological space, then everything we have proven about that kind of topological space can be automatically applied to that object. This is relevant because it turns out that a number of objects one studies in mathematics are topological spaces in disguise (with some overlying structure) such as your metric spaces from analysis. Cf. my post here along these lines; the motivation for why the topological axioms have been chosen as such is often overlooked at least at the start of the intro course, and I feel that if the motivation were properly addressed from the start, everything would become clear faster.
The idea of a subspace topology provides the rigorous framework upon which statements like this one from calculus can be made:

If a function $f:\mathbb{R} \to \mathbb{R}$ is continuous, then so too is the restriction of that function to a particular interval, say to $[0,1]$.

Here, it is implied that $[0,1]$ is inheriting the subspace topology from $\mathbb{R}$, which is endowed with the so-called Euclidean topology.
A: To talk about a topological space, you need a set and a topology on that set.
If you are given a topological space $X$, you're actually given a pair $(X, \mathcal T_X).$
Now, if you're given a subset $A \subseteq X$, you want to get a topological space out of it. To say that "$A$ is equipped with the subspace topology", we mean that we're looking at the topological space $(A, \mathcal T_A)$, where $\mathcal T_A$ is as you described.
Note that this mainly stems from the abuse of notation we do when we say something like "Let $X$ be a topological space", instead of using $(X, \mathcal T_X)$.
A: Nothing is altered about the set $A$ itself. Instead, the language "to equip with a topology" means to spawn a topological space $(A, T_A)$ with the topology $T_A$ being the topology in question. The topological space $(A, T_A)$ and the set $A$ are not the same thing. $A$ forms a part of it, but it is not identical to $A$, nor does it alter the definition of $A$.
How much differentiation there is between these two things depends on how rich our formalizing foundations are. With ordinary, "ZFC" set theory, $(A,\ T_A)$ is just an ordered pair of sets, which in turn is the set $\{ \{ A\}, \{ A, T_A \} \}$. Note this is not equal to $A$, nor does it have $A$ as a direct member. A richer system of foundations, though, would incorporate a typed language and say that $(A, T_A)$ is an object of a different "data type", the type of topological spaces, and thus could not be interconverted to a set or other type of ordered pair unless explicit type casting rules are defined.
