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Could someone explain the intuition behund the Hausdorff Measure and Hausdorff Dimension?
The Hausdorff Measure is defined as the following:

Let $(X,d)$ be a metric space. $\forall S \subset X$, let $\operatorname{diam} U$ denote the diameter, that is $$\operatorname{diam} U = \sup \{ \rho(x,y) : x,y \in U \} \,\,\,\,\, \operatorname{diam} \emptyset = 0 $$ Let $S$ be any subset of $X$ and $\delta > 0 $ a real number. Define $$ H^{d}_{\delta} (S) := \inf \{ \sum_{i=1}^{\infty} (\operatorname{diam} U_i )^d : \bigcup_{i=1}^{\infty} U_i \supseteq S, \operatorname{diam} U_i < \delta \}$$

What does $\rho(x,y)$ denote? What is meant by the diameter of a set? I'm just trying to understand the intuition behind this definition.
The Hausdorff Dimension is defined as the following:

Let $X$ be a metric space. If $S \subset X$ and $d \in \mathbb{R}^+$, the Hausdorff content is defined as $$C^{d}_H (S) := \inf \{ \sum_{i} r_i^d : \, \, \text{there is a cover of $S$ by balls of radii} \, \, r_i > 0 \} $$ Then the Hausdorff Dimension is defined as
$$\operatorname{dim}_{H} (X) := \inf \{d \geq 0 : C^d_H = 0\}$$

Could someone explain the intuition behind these definitions?

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What does $\rho (x,y)$ denote?

It denotes $d(x,y)$. That is, you have a typo somewhere: $d$ and $\rho$ are the same thing, the metric.

What is meant by the diameter of a set?

What you wrote: the supremum of pairwise distances between the points in your set. To develop intuition, draw a few shapes on the plane (which is an excellent example of a metric space) and determine their diameters. The diameter of a circle is just that, the diameter (hence the name). The diameter of a triangle is the length of its longest side. And so on.

I'm just trying to understand the intuition behind this definition.

The intuition is that if the object is $d$-dimensional, then $r^d$ roughly represents the volume of its piece of size $r$. Summing over all pieces, we should get something that is no less than the volume of the object. That is, the sums should not be arbitrarily close to zero. And if they are, then the value of $d$ we picked is too high, and the actual dimension of the object is lower than that. So we make $d$ smaller (i.e., take infimum over $d$).

The preceding paragraph is a lie, but this is what you get when you ask for intuition.

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Let me try to explain, the way I understand, the intuition behind the definition of Hausdorff dimension. As user103254 said, $\rho =d$. Now let us look at some examples.

As always, intuition is derived from $\mathbb{R}$, $\mathbb{R}^2$ and $\mathbb{R}^3$. For simplicity, I would use cubes of sides $r_i$ rather than balls. A cube in $\mathbb{R}$ is an interval, a cube in $\mathbb{R}^2$ is a square, a cube in $\mathbb{R}^3$ is the usual cube and so on. Also, I would assume that all cubes are of same side-length. Thus the sum of the radii is nothing but the number of cubes multiplied by the side-length. First look at the interval [0,1]. If we want to cover it with intervals of size $\frac{1}{n}$ the best we can hope to do is to cover it with $n$ closed intervals of length $\frac{1}{n}$. Similarly you need $n^2$ squares of sides $\frac{1}{n}$ to cover $[0,1]\times [0,1]$, $n^3$ cubes of sides $\frac{1}{n}$ to cover $[0,1]\times [0,1]\times [0,1]$ and so on. This can also be seen from the fact that the volume of a $d$-dimensional cube of side $\frac{1}{n}$ is intuitively $\frac{1}{n^d}$ where as the volume of $[0,1]^d$ is $1$. If the size of the box is fixed, then higher the dimension the more boxes you would require to cover.

Thus, the sum of radii in the definition of $d$-dimensional content of $[0,1]^k$ in this special case will look like $n^k \times \frac{1}{n^d} = n^{k-d}$. As $n$ tends to infinity, $n^{d-k}$ tend to zero if $d \geq k$ and infinity otherwise. So naively the dimension of $[0,1]^k$ is $k$. I say naively because while calculating the infinum we could have taken covers of different radii. Of course we did not even take balls we took cubes. But I guess the idea is clear.

This I think is the basic intuition. Hope it was helpful.

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