I love being inspired!

$x\quad =\quad \sqrt { 2\quad +\quad \frac { 2 }{ \sqrt { 2\quad -\quad \frac { 2 }{ \sqrt { 2\quad +\frac { 2 }{ \sqrt { \dots } } } } } } } \\ \\ Squaring\quad Both\quad Sides\\ \\ { x }^{ 2 }\quad =\quad 2\quad +\quad \frac { 2 }{ \sqrt { 2\quad -\quad \frac { 2 }{ \sqrt { 2\quad +\frac { 2 }{ \sqrt { \dots } } } } } } \\ \Longrightarrow \quad { x }^{ 2 }\quad =\quad 2\quad +\quad \frac { 2 }{ \sqrt { 2-\quad \frac { 2 }{ x } } } \\ \\ \Longrightarrow \quad { (x }^{ 2 }-2)^{ 2 }\quad =\quad \frac { 4 }{ 2\quad -\quad \frac { 2 }{ x } } \\ { x }^{ 4 }+4-4{ x }^{ 2 }\quad =\quad \frac { 2 }{ 1\quad -\quad \frac { 1 }{ x } } \Longrightarrow \quad (1-\frac { 1 }{ x } )({ x }^{ 4 }+4-4{ x }^{ 2 })\quad =\quad 2\\ { x }^{ 5 }-{ x }^{ 4 }-4{ x }^{ 3 }+4{ x }^{ 2 }+2x\quad =4\\ \\ We\quad can\quad see\quad that\quad if\quad we\quad put\quad anything\quad in\quad \\ the\quad x\quad in\quad { x }^{ 5 }-{ x }^{ 4 }-4{ x }^{ 3 }+4{ x }^{ 2 }\quad we\quad will\quad get\quad 0\\ \\ Hence\quad we\quad could\quad see\quad that\quad only\quad term\quad left\quad is\quad 2x\\ \\ Therefore\quad x\quad =\quad 2\quad and\quad 2(x)\quad =\quad 4\\$

$\color{#3D99F6} { 1000*2 = \boxed{2000} }$

$\huge \quad \quad \color{#D61F06}{Alternative}$

$We\quad may\quad also\quad notice\quad that:\\ \\ { x }^{ 5 }-{ x }^{ 4 }-4{ x }^{ 3 }+4{ x }^{ 2 }+2x\quad =4\\ \\ { x }^{ 5 }-{ x }^{ 4 }-4{ x }^{ 3 }+4{ x }^{ 2 }+2x-4\quad =\quad 0\\ \\ { x }^{ 5 }-{ x }^{ 4 }-{ x }^{ 3 }\quad \quad -{ x }^{ 3 }+{ x }^{ 2 }+x\quad \quad -{ x }^{ 3 }+{ x }^{ 2 }+x\quad \quad -{ x }^{ 3 }+{ x }^{ 2 }+{ x }^{ 2 }\quad -4\quad =\quad 0\\ \\ { x }^{ 3 }({ x }^{ 2 }-x-1)\quad \quad -x({ x }^{ 2 }-x-1)\quad \quad -x({ x }^{ 2 }-x-1)\quad -{ x }^{ 3 }+{ x }^{ 2 }+{ x }^{ 2 }\quad -4\quad =\quad 0\\ \\ ({ x }^{ 3 }-2x)({ x }^{ 2 }-x-1)\quad \quad -{ x }^{ 3 }+2{ x }^{ 2 }\quad -4\quad =\quad 0\\ \\ \\ ({ x }^{ 3 }-2x)\quad =\quad { x }^{ 3 }-2{ x }^{ 2 }\quad +4\quad \Longrightarrow \quad { x }^{ 2 }-x-2\quad =\quad 0\\ \Longrightarrow (x-2)(x+1)\quad =\quad 0\quad \Longrightarrow \quad x\quad =\quad 2\quad (Only)$

$\huge \color{#20A900}{Alternative}$

$Let\quad y\quad be\quad similar\quad equation:\\ \\ y\quad =\quad \sqrt { 2\quad -\quad \frac { 2 }{ x } } and\quad x\quad =\quad \sqrt { 2\quad +\quad \frac { 2 }{ y } } \\ { x }^{ 2 }-{ y }^{ 2 }\quad =\quad \frac { 2 }{ x } +\frac { 2 }{ y } \Longrightarrow (x+y)(x-y)=\frac { x+y }{ xy } \Rightarrow x-y=\frac { 2 }{ xy } \\ x+y\quad \ncong \quad 0\quad \\ (x-y)xy\quad =\quad 2\Rightarrow (x-y)\quad =\quad 2\\ x(x-2)=2\\$ And Complete further . . . .

x = $\boxed {2}$

Let $y = \sqrt{2-\frac 2{\sqrt{2+\frac 2{\sqrt{2-\frac 2{\sqrt \cdots}}}}}}$ . Then we note that:

$\begin{cases} x = \sqrt{2+\dfrac 2y} & \implies x^2 = 2 + \dfrac 2y & ...(1) \\ y = \sqrt{2-\dfrac 2x} & \implies y^2 = 2 - \dfrac 2x & ...(2) \end{cases}$

$\begin{aligned}(1) - (2): \quad x^2 - y^2 & = \frac 2y + \frac 2x \\ (x-y)(x+y) & = \frac {2(x+y)}{xy} \\ \implies x-y & = \frac 2{xy} & ...(3) \end{aligned}$

$\begin{aligned}(1) + (2): \quad x^2 + y^2 & = 4 + \frac 2y - \frac 2x \\ & = 4 + \color{#3D99F6} \frac {2(x-y)}{xy} & \small \color{#3D99F6} (3): \ \frac 2{xy} = x-y \\ & = 4 + \color{#3D99F6} (x-y)^2 \\ x^2 + y^2 & = 4 + x^2-2xy + y^2 \\ \implies xy & = 2 \implies x = \frac 2y \end{aligned}$

Therefore,

$\begin{aligned} (1): \quad x^2 & = 2 + \frac 2y \\ & = 2 + x \\ \implies x^2 - x - 2 & = 0 \\ (x-2)(x+1) & = 0 \\ \implies x & = 2 & \small \color{#3D99F6} \because x > 0 \end{aligned}$

Then $1000x = \boxed{2000}$ .