Prove that the leverage converges to 0. Do the residuals in a linear regression really approximate the errors? In linear regression, $y_i = x_i^T \beta + \epsilon_i$, $i=1,\dots,n$, and $Var(\epsilon_i)=\sigma^2$. It is well known that the residuals $e_i$ have variance $Var(e_i) = \sigma^2 (1-h_{ii})$, where $h_{ii}$ is the leverage of observation $i$, and that is not equal to $\sigma^2$.
It is commonly believed that the leverage goes to 0, and the residuals $e_i$ approximate the errors $\epsilon_i$. But how can this be justified?
Clearly, it is necessary that the leverage converges to 0 in order for the residuals to approximate the errors, so that they have the same variance asymptotically. But I can find no reference that proves: for all $i=1,\dots,n$, $h_{ii}\to0$ as $n\to\infty$.
It is known that the average of the leverage is $p/n$, where $p$ is the number of parameters, and $h_{ii}>0$. But the mean of a sequence of positive numbers converging to 0, does not imply the elements of the sequence converge to 0. In fact, $h_{ii}$ depends on $x_i$, so it would seem any proof would need to place some assumption on how $x_i$, $i=1,\dots,n$, grows for large $n$.
How does one prove that leverage converges to 0? Is there a reference?

Your conjecture is not true. It is probably true when the design matrix $X$ satisfies some regularity conditions, but in general, there is no guarantee that every $h_{ii}$ converges to $0$ (technically, when $X$ is not deterministic, you should also clarify that the convergence mode of $h_{ii}$, in probability or almost surely?).
For example, let $X$ be (it's a deterministic matrix but depends on the sample size $n$):
\begin{align}
X = \begin{bmatrix}
1 & n \\
1 & 0 \\
\vdots & \vdots \\
1 & 0 
\end{bmatrix} \in \mathbb{R}^{n \times 2}. 
\end{align}
It is then easy to verify that $h_{11} = 1$, $h_{ii} = (n - 1)^{-1}$ for $2 \leq i \leq n$. Therefore $h_{11} \not\to 0$ as $n \to \infty$.
