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: 
write $$\sqrt{n+1}\le 1+\sqrt{n}$$ after squaring (all summands are positive) we get
$$0\le \sqrt{n}$$ which is true.
Another method:

Show that $\sqrt{x+1}-\sqrt{x} \le 1$ for all non-negative real numbers ... which of course includes all the non-negative integers.

One way to do this is to show that the function $f(x)=\sqrt{x+1}-\sqrt{x}$ equals $1$ for $x=0$ and that for all $x$: $f'(x) <0$, i.e. $f(x)$ is a monotonically decreasing function.