Using standard techniques for finding closed-form solutions to recurrences, one can show that $F_n=\frac{\varphi^n-\hat\varphi^n}{\sqrt5}=\frac{\varphi^n}{\sqrt5}-\frac{\hat\varphi^n}{\sqrt5}\;,\tag{1}$ where $\varphi=\frac12(1+\sqrt5)$ and $\hat\varphi=\frac12(1-\sqrt5)$. Now $\left|\frac{\hat\varphi}{\sqrt5}\right|\approx -0.2764<\frac12\;,$ and $|\hat\varphi|\approx 0.618<1$, so $\left|\frac{\hat\varphi^n}{\sqrt5}\right|<\frac12$ for all $n\ge 0$.
Thus, $\left|F_n-\frac{\varphi^n}{\sqrt5}\right|=\left|\frac{\hat\varphi^n}{\sqrt5}\right|<\frac12$ for all $n\ge 0$. But we know that $F_n$ is an integer, so $F_n$ is an integer less than half a unit away from $\dfrac{\varphi^n}{\sqrt5}$. That means that $F_n$ is the unique integer closest to $\dfrac{\varphi^n}{\sqrt5}$, so that rounding $\dfrac{\varphi^n}{\sqrt5}$ to the nearest integer gives you $F_n$. In particular,
$F_n=\left\lfloor\frac{\varphi^n}{\sqrt5}+\frac12\right\rfloor\;,$ where $\lfloor x\rfloor$ is the largest integer less than or equal to $x$. However, I’m not actually going to use that result; the argument that you posted is slightly defective, and it’s easier to do it right using $(1)$.
A number $m$ has $1000$ digits if and only if $10^{999}\le m<10^{1000}$: $10^{999}$ is the first integer with $1000$ digits, and $10^{1000}-1$ is the last. Thus, we need the smallest $n$ such that $F_n\ge 10^{999}$, or, using $(1)$, the smallest $n$ such that
$\frac{\varphi^n}{\sqrt5}-\frac{\hat\varphi^n}{\sqrt5}\ge 10^{999}\;.$
Clearly $n$ is going to be quite large, and we saw earlier that the second term on the left is going to be very small, so to a good first approximation we want the smallest $n$ such that $\frac{\varphi^n}{\sqrt5}\ge 10^{999}\;;\tag{2}$ we’ll find that and then come back to make sure that the second term is too small to matter.
$(2)$ is equivalent to $\varphi^n\ge 10^{999}\sqrt5$ and hence, after taking logarithms, to $n\ln\varphi\ge 999\ln 10+\ln\sqrt5=999\ln 10+\frac12\ln 5\;.\tag{3}$
Dividing both sides of $(3)$ by $\ln\varphi$, we get
$n\ge\frac{999\ln 10+\frac12\ln 5}{\ln\varphi}\approx 4781.86\;,$
where the numerical approximation is taken from a calculator. Clearly the smallest integer $n\ge 4781.86$ is $4782$. A little more work with a calculator shows that $\log_{10}\frac{\varphi^{4782}}{\sqrt5}\approx 999.0294106732$ and hence that $\frac{\varphi^{4782}}{\sqrt5}\approx 1.07\times 10^{999}\;.$
Since $\dfrac{\hat\varphi^{4782}}{\sqrt5}$ is clearly much less than $7\times 10^{997}$, $F_{4782}\ge 10^{999}$, and our approximation in $(2)$ did no harm.
As an extra check, note that $\frac{\varphi^{4781}}{\sqrt5}<7\times 10^{998}\;,$ and $\dfrac{\hat\varphi^{4781}}{\sqrt5}$ is clearly much less than $3\times 10^{998}$, so $F_{4781}<10^{999}$ and has only $999$ digits.