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$\newcommand\de\delta$The answer is no. In fact, $d_H(A,C_n(L_n))$ can be however large or even infinite for all $n$.

Indeed, e.g. suppose that $X=[-1,a)$ for some $a\in(0,\infty]$ endowed with the standard metric $|\cdot-\cdot|$, $\ell(x)=u(x)=-x$ for $x\in[-1,0]$, $\ell(x)=u(x)=0$ for $x\in[0,a)$. Let $(L_n)$ be any positive sequence converging to $0$, and let $p_n=2L_n$ eventually (that is, for all large enough $n$). Then $A\supseteq[0,a)$ and $C_n(L_n)=[-3L_n,-L_n]$ eventually, so that eventually $$d_H(A,C_n(L_n))\ge|(a-L_n)-(-L_n)|=a$$ if $a<\infty$, and $d_H(A,C_n(L_n))=\infty$ if $a=\infty$.

Details: Let $A_n(\de):=\{x\in X\colon d(p_n,[\ell(x),u(x)])\le\de\}$ and $A(\de):=\liminf_{n\to\infty}A_n(\de)$, so that $A=\bigcap_{\de>0}A(\de)$.

For each $\de>0$, eventually $A_n(\de)\supseteq[0,a)$, so that $A(\de)\supseteq[0,a)$ for all $\de>0$ and hence $A\supseteq[0,a)$, as was claimed.

Also, $C_n(L_n)=A_n(L_n)=\{x\in[-1,0]\colon |p_n-(-x)|\le L_n\}=[-3L_n,-L_n]$, as was claimed, too.

$\newcommand\de\delta$The answer is no. In fact, $d_H(A,C_n(L_n))$ can be however large or even infinite for all $n$.

Indeed, e.g. suppose that $X=[-1,a)$ for some $a\in(0,\infty]$ endowed with the standard metric $|\cdot-\cdot|$, $\ell(x)=u(x)=-x$ for $x\in[-1,0]$, $\ell(x)=u(x)=0$ for $x\in[0,a)$. Let $(L_n)$ be any positive sequence converging to $0$, and let $p_n=2L_n$ eventually (that is, for all large enough $n$). Then $A\supseteq[0,a)$ and $C_n(L_n)=[-3L_n,-L_n]$ eventually, so that eventually $$d_H(A,C_n(L_n))\ge|(a-L_n)-(-L_n)|=a$$ if $a<\infty$, and $d_H(A,C_n(L_n))=\infty$ if $a=\infty$.

Details: Let $A_n(\de):=\{x\in X\colon d(p_n,[\ell(x),u(x)])\le\de\}$ and $A(\de):=\liminf_{n\to\infty}A_n(\de)$, so that $A=\bigcap_{\de>0}A(\de)$.

For each $\de>0$, eventually $A_n(\de)\supseteq[0,a)$, so that $A(\de)\supseteq[0,a)$ for all $\de>0$ and hence $A\supseteq[0,a)$, as was claimed.

Also, $C_n(L_n)=A_n(L_n)=\{x\in[-1,0]\colon |p_n-(-x)|\le L_n\}=[-3L_n,-L_n]$ eventually, as was claimed, too.

$\newcommand\de\delta$The answer is no. In fact, $d_H(A,C_n(L_n))$ can be however large or even infinite for all $n$.

Indeed, e.g. suppose that $X=[-1,a)$ for some $a\in(\frac14,\infty]$ endowed with the standard metric $|\cdot-\cdot|$, $\ell(x)=u(x)=-x$ for $x\in[-1,0]$, $\ell(x)=u(x)=0$ for $x\in[0,a)$, $p_n=\frac1{n+1}$, and $L_n=\frac1{2n+2}$. Then $A\supseteq[0,a)$ and $C_n(L_n)=[-\frac3{2n+2},-\frac1{2n+2}]$, so that $$d_H(A,C_n(L_n))\ge\Big|\Big(a-\frac1{2n+2}\Big)-\Big(-\frac1{2n+2}\Big)\Big|=a$$ if $a<\infty$, and $d_H(A,C_n(L_n))=\infty$ if $a=\infty$.

Details: Let $A_n(\de):=\{x\in X\colon d(p_n,[\ell(x),u(x)])\le\de\}$ and $A(\de):=\liminf_{n\to\infty}A_n(\de)$, so that $A=\bigcap_{\de>0}A(\de)$.

Note that $A_n(\de)\supseteq[0,a)$ for all $n\ge1/\de$, so that $A(\de)\supseteq[0,a)$ for all $\de>0$ and hence $A\supseteq[0,a)$, as was claimed.

Also, $C_n(L_n)=A_n(L_n)=\{x\in[-1,0]\colon |\frac1{n+1}-(-x)|\le\frac1{2n+2}\}=[-\frac3{2n+2},-\frac1{2n+2}]$, as was claimed, too.

$\newcommand\de\delta$The answer is no. In fact, $d_H(A,C_n(L_n))$ can be however large or even infinite for all $n$.

Indeed, e.g. suppose that $X=[-1,a)$ for some $a\in(0,\infty]$ endowed with the standard metric $|\cdot-\cdot|$, $\ell(x)=u(x)=-x$ for $x\in[-1,0]$, $\ell(x)=u(x)=0$ for $x\in[0,a)$. Let $(L_n)$ be any positive sequence converging to $0$, and let $p_n=2L_n$ eventually (that is, for all large enough $n$). Then $A\supseteq[0,a)$ and $C_n(L_n)=[-3L_n,-L_n]$ eventually, so that eventually $$d_H(A,C_n(L_n))\ge|(a-L_n)-(-L_n)|=a$$ if $a<\infty$, and $d_H(A,C_n(L_n))=\infty$ if $a=\infty$.

Details: Let $A_n(\de):=\{x\in X\colon d(p_n,[\ell(x),u(x)])\le\de\}$ and $A(\de):=\liminf_{n\to\infty}A_n(\de)$, so that $A=\bigcap_{\de>0}A(\de)$.

For each $\de>0$, eventually $A_n(\de)\supseteq[0,a)$, so that $A(\de)\supseteq[0,a)$ for all $\de>0$ and hence $A\supseteq[0,a)$, as was claimed.

Also, $C_n(L_n)=A_n(L_n)=\{x\in[-1,0]\colon |p_n-(-x)|\le L_n\}=[-3L_n,-L_n]$, as was claimed, too.

The answer is no. In fact, $d_H(A,C_n(L_n))$ can be however large or even infinite for all $n$.

Indeed, e.g. suppose that $X=[-1,a)$ for some $a\in(\frac14,\infty]$, $\ell(x)=u(x)=-x$ for $x\in[-1,0]$, $\ell(x)=u(x)=0$ for $x\in[0,a)$, $p_n=\frac1{n+1}$, and $L_n=\frac1{2n+2}$. Then $A\supseteq[0,a)$ and $C_n(L_n)=[-\frac3{2n+2},-\frac1{2n+2}]$, so that $$d_H(A,C_n(L_n))\ge d\Big(a-\frac1{2n+2},-\frac1{2n+2}\Big)=a$$ if $a<\infty$, and $d_H(A,C_n(L_n))=\infty$ if $a=\infty$.

$\newcommand\de\delta$The answer is no. In fact, $d_H(A,C_n(L_n))$ can be however large or even infinite for all $n$.

Indeed, e.g. suppose that $X=[-1,a)$ for some $a\in(\frac14,\infty]$ endowed with the standard metric $|\cdot-\cdot|$, $\ell(x)=u(x)=-x$ for $x\in[-1,0]$, $\ell(x)=u(x)=0$ for $x\in[0,a)$, $p_n=\frac1{n+1}$, and $L_n=\frac1{2n+2}$. Then $A\supseteq[0,a)$ and $C_n(L_n)=[-\frac3{2n+2},-\frac1{2n+2}]$, so that $$d_H(A,C_n(L_n))\ge\Big|\Big(a-\frac1{2n+2}\Big)-\Big(-\frac1{2n+2}\Big)\Big|=a$$ if $a<\infty$, and $d_H(A,C_n(L_n))=\infty$ if $a=\infty$.

Details: Let $A_n(\de):=\{x\in X\colon d(p_n,[\ell(x),u(x)])\le\de\}$ and $A(\de):=\liminf_{n\to\infty}A_n(\de)$, so that $A=\bigcap_{\de>0}A(\de)$.

Note that $A_n(\de)\supseteq[0,a)$ for all $n\ge1/\de$, so that $A(\de)\supseteq[0,a)$ for all $\de>0$ and hence $A\supseteq[0,a)$, as was claimed.

Also, $C_n(L_n)=A_n(L_n)=\{x\in[-1,0]\colon |\frac1{n+1}-(-x)|\le\frac1{2n+2}\}=[-\frac3{2n+2},-\frac1{2n+2}]$, as was claimed, too.