Question: I have some difficulties proving the following statement: $\sqrt {n+1} - \sqrt n ≤ 1$

Thus far I have the following:

The root of a positive integer is always smaller than or equal to the integer, since $a^2 ≥ a$ where a is a positive integer.

Therefore $\sqrt {n+1} ≤ \sqrt n + 1$ and thus $\sqrt {n+1} - \sqrt n ≤ 1$

However I feel like I am jumping to a conclusion and I'd like to be more precise.

This form would be precise if I'd want to prove any of the following:

$\sqrt {n+1} ≤ n + 1$

or
$\sqrt n + \sqrt 1 ≤ \sqrt n + 1$

But I don't think it is precise enough for what I actually want to prove. Any ideas?
I understand the statement is true. Adding an integer $x$ to an integer $n$ and then taking the square root of that will always be less than the squre root of $n$ + $x$, I just don't know how to prove it.

Answer: 
Hint:

$$\sqrt{n+1}-\sqrt{n}=\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}}$$

Use $(a+b)(a-b)=(a^2-b^2)$ for the numerator.
Another way,
which does not use the standard
$a^2-b^2
= (a-b)(a+b)$:

Since
$(\sqrt{n}+1/2)^2
= n+\sqrt{n}+\frac14
\gt n+1
$
for $n \ge 1$,
$\sqrt{n+1}
\lt \sqrt{n}+\frac12$.