For me you proof is very confusing and it is wrong. It is wrong because the equation $|f(x)-f(2)| < \frac{\delta}{\sqrt{3}+\sqrt{2}}$ does not hold. $|f(2.1)-f(2)|\approx 0.034$, but $\frac{0.1}{\sqrt3+\sqrt2}\approx 0.032$.
But now to the confusion. You state
Givens: $\varepsilon > 0$ and $\delta > 0$ and $|x-2| < \delta$ for $\delta < 1$.
and then you state
Let $0 < \delta < \min{\{1,(\sqrt{3}+\sqrt{2})\varepsilon}\}$
But what is the difference? The inequalities $0 < \delta < \min{\{1,(\sqrt{3}+\sqrt{2})\varepsilon}\}$ is also given. And if $\delta < \min{\{1,(\sqrt{3}+\sqrt{2})\varepsilon}\}$ then $\delta < 1$ is redundant. So you can leave out this statement. Finally you do not have to prove that there is an interval of $\delta$ where your statment holds, it is sufficient that for each $\varepsilon$ you find on $\delta$. From you inequalities I deduce that $\min \{\frac12, \varepsilon\}$ is such a $\delta$, so I would write
Given $\varepsilon>0$ choose $\delta=\min{\frac12, \varepsilon}$ and assume that $|x-2|<\delta$.
Then you state:
$$|f(x)-f(2)| = |\sqrt{x}-\sqrt{2}| < \varepsilon \tag1$$
$$ x-2 < |\sqrt{x}+\sqrt{2}|\varepsilon \tag2$$
And what's that? I think you want to tell us
We have to show: $$|f(x)-f(2)| = |\sqrt{x}-\sqrt{2}| < \varepsilon \tag1$$
I think you want to tell us that and not that this is a prerequisite.
So I think you should state this explicitly.
An what is $(2)$? We have
$$(\sqrt x-\sqrt 2)(\sqrt x+\sqrt 2)=(x-2)$$
This is formula is crucial to the proof so I would explicitly mention it explicitly.
From this and $(1)$ follows $$ x-2 < |\sqrt{x}+\sqrt{2}|\varepsilon \tag2$$
$(3)$ and $(4)$ follow immediately from the definition of $\delta$ and the $|x-2|<\delta$.
I do not understand why you write "Then it follows" to justify $(4)$. Because it does not follow from $3$
Finally you do not tell us how you get inequality $(5)$. As I mentioned at the beginning. This inequality is wrong.
Here is a short proof:
Assume that $\varepsilon>0$ and $\delta=\varepsilon.$ If $|x-2|<\delta$ then
$$|f(x)-f(2)|=|\sqrt x-\sqrt 2|=\frac{|\sqrt x-\sqrt 2|\cdot|\sqrt x+\sqrt 2|}{|\sqrt x+\sqrt 2|}=\frac{|x-2|}{\sqrt x+\sqrt 2}\lt\frac{\delta}{\sqrt 2}<\varepsilon$$
So $f(x)=\sqrt x$ is continuous at $x=2.$