My Proof:

For convenience and clarity, let’s define some real functions for the arguments of the floor function. We’ll limit the domains (inputs) to positive values to simplify later steps.

\begin{align*}
f(x)&:= \sqrt{x}+\sqrt{x+1}, &x>0 \\
g(x)&:=\sqrt{4x+2}, &x>0
\end{align*}

As x>0, it is easy to see that: f(x) > 0; f'(x)>0; g(x)>0; g'(x)>0. That is, both f(x) and g(x) are positive and increasing with increasing x.

**Lemma 1:** f(x)<g(x) for all x>0.

As both functions are positive, we can compare them by comparing their squares.

\begin{align*}
f^2(x)=(\sqrt{x}+\sqrt{x+1})^2 = x + 2\sqrt{x^2+x}+x+1=&2x+1+\sqrt{4x^2+4x} \\
g^2(x)=4x+2=2x+1+(2x+1)=&2x+1+\sqrt{4x^2+4x+1}
\end{align*}

Looking at these it is clear that f^2(x)<g^2(x), therefore f(x)<g(x).

Inspecting tabulated values for \lfloor{f(n)}\rfloor and \lfloor{g(n)}\rfloor (thanks @gerthworm!), it appears that these are indeed equal, and increase for numbers which can be expressed as m^2 or m^2+m, m\in{\mathbb{N}}.

Based on this, we define m:=\lfloor\sqrt{n}\rfloor.

**Lemma 2:** m^2\leq n\leq m^2+m.

As m=\lfloor\sqrt{n}\rfloor\leq\sqrt{n}, m^2\leq{n}. Also, n< m^2+2m+1 because

n\geq m^2+2m+1\implies\sqrt{n}\geq m+1\implies \sqrt{n}\geq \lfloor{\sqrt{n}}\rfloor+1

which clearly false. But this implies n\leq m^2+m, proving both parts of the lemma.

**Conjecture, from tabulation:**

\begin{align*}
m^2\leq &n \leq m^2+m-1:&&\lfloor{f(n)}\rfloor=\lfloor{g(n)}\rfloor=2m \\
m^2+m\leq &n \leq m^2+2m:&&\lfloor{f(n)}\rfloor=\lfloor{g(n)}\rfloor=2m+1
\end{align*}

From Lemma 2, these two cases combine to include all n\in\mathbb{N}.

As we have already shown that f(x) and g(x) are increasing with increasing x, and that f(x)<g(x), we need to prove two things for each half of the conjecture, in particular:

\begin{align*}
&1: &2m&\leq f(m^2) \\
&2: &g(m^2+m-1)&<2m+1 \\
&3: &2m+1&\leq f(m^2+m) \\
&4: &g(m^2+2m)&<2m+2
\end{align*}

Actually, let’s go a little better and prove them all as strict inequalities. In case 3, we use that both f(x) and 2m+1 are positive to take the square root of the square.

\begin{align*}
1: f(m^2)&=\sqrt{m^2}+\sqrt{m^2+1} \\
&>\sqrt{m^2}+\sqrt{m^2} \\
&>2m
\end{align*}

\begin{align*}
2: g(m^2+m-1)&=\sqrt{4m^2+4m-4+2} \\
&<\sqrt{4m^2+4m+1} \\
&<2m+1
\end{align*}

\begin{align*}
3: f(m^2+m)&=\sqrt{m^2+m}+\sqrt{m^2+m+1} \\
&=\sqrt{(\sqrt{m^2+m}+\sqrt{m^2+m+1})^2} \\
&=\sqrt{m^2+m +2\sqrt{m^2+m}\sqrt{m^2+m+1}+m^2+m+1}\\
&=\sqrt{2m^2+2m+1 +2\sqrt{m^2+m}\sqrt{m^2+m+1}}\\
&>\sqrt{2m^2+2m+ 1+2\sqrt{m^2+m}\sqrt{m^2+m}}\\
&>\sqrt{2m^2+2m+ 1+2(m^2+m)}\\
&>\sqrt{4m^2+4m+ 1}\\
&>\sqrt{(2m+1)^2}\\
&>2m+1
\end{align*}

\begin{align*}
4: g(m^2+2m)&=\sqrt{4m^2+8m+2}\\
&<\sqrt{4m^2+8m+4}\\
&<2m+2
\end{align*}

As we have proven the conjecture, we have also proven the goal, because the two parts combine to prove \lfloor{f(n)}\rfloor=\lfloor{g(n)}\rfloor for every natural number.