How to prove $\frac{\partial}{\partial x^r} (x^r) = 1$ where $(U, (x^1, ..., x^n))$ is a local chart of a smooth manifold.

Good question, these basic details are really important and worth thinking about once in a while, in my opinion.

I guess the answer really depends on how you define the tangent space to begin with, though, so I'll give you my version. In the introductory course to Riemannian Geometry that I took, given $p\in M$, we defined $T_pM$ as follows:

Let $c:(-\varepsilon,\varepsilon)\to M$ be a smooth curve on M such that $c(0)=p$. Consider the set $C^\infty(p)$ of all functions $f: M\to\mathbb{R}$ that are differentiable at $p$. The $\textbf{tangent vector to the curve}$ $c$ at $p$ is then the operator $\dot{c}(0): C^\infty(p)\to\mathbb{R}$ given by

$$\dot{c}(0)(f)=\frac{d(f\circ c)}{dt}(0)$$

And then given a neighborhood $U\subset M$ of $p$ and a local chart $\varphi=(x^1,\ldots,x^n): U\to\mathbb{R}^n$, we define $\left(\frac{\partial}{\partial x_i}\right)_p$ as just notation for the vector tangent to the curve $c:(-\varepsilon,\varepsilon)\to M$ given in coordinates by:

$$\left(\varphi\circ c\right)(t)=(x^1_p,\ldots,x^i_p+t,\ldots,x^n_p)$$

where $(x^1_p,\ldots,x^n_p)=\varphi(p)$. So, having defined everything, we can compute what you asked for:

$$\left(\frac{\partial}{\partial x_i}\right)_p(x^i)=\frac{d(x^i\circ c)}{dt}(0)=\frac{d(x^i_p+t)}{dt}(0)=1$$.

You can then prove that $T_pM$ is indeed a vector space, for which the $\left(\frac{\partial}{\partial x_i}\right)_p$'s defined in this way form a basis.

Thank you for the question. I hope this helps.

You just have to know what $\frac{\partial}{\partial x^r}$ is, $r \in \{1 , \ldots , n\}$. At first you have to consider a chart $(U , \psi = (x^1 , \ldots , x^n))$ of your manifold $M$. Well, if $p \in U$, you have to know that there exists a biyection $h_{(U , \psi)} : T_pM \to {\mathbb{R}}^n$, which can be used to show that $T_pM$ is a $n$-dimensional vector space (that map is therefore an isomorphism). By notation, you type now $$ {\left(\frac{\partial}{\partial x^r}\right)}_p := h_{(U , \psi)}^{- 1}(e_r)\mbox{,} $$ where ${\{e_r\}}_{r = 1}^n$ is the canonical base of ${\mathbb{R}}^n$.