Proving $ \forall r, s \in \mathbb{R^+} \sqrt {r. s} = \sqrt r . \sqrt s$ I am new to writing proofs, I have written down a proof for $ \forall r, s \in \mathbb{R^+}  \sqrt {r\cdot s}  = \sqrt r \cdot  \sqrt s$
I know I am a bit messy with my proof, I would love to get some feedback. Also please verify if my proof is good enough. Here is my approach,
Proof -  Let, $\sqrt r = p$ and $\sqrt s = q$ where  $p, q \in \mathbb{R^+}$.
Then,
$\sqrt r \sqrt s  = p\cdot q$ ... (1)
Squaring both the sides we get,
$(\sqrt r \sqrt s)^2  = (p\cdot q)^2$   ...(2)
$(\sqrt r \sqrt s)^2 = (r^{1/2}\cdot s^{1/2})^2$  [as $\sqrt a = a^{1/2}$]

Lemma 1 - $\forall a, b \in  \mathbb{R} [(a\cdot  b )^2 = a^2\cdot  b^2] $
Proof - By definition of squaring, $ a^2 = a\cdot a$
It follows from the definition that, $(a \cdot b)^2 = (a \cdot b ) (a\cdot b) = a^2\cdot b^2$ $\blacksquare$

From lemma 1,
$(r^{1/2}\cdot s^{1/2})^2 = r^{2 / 2}\cdot  s^{2/2} = r\cdot s$
Therefore,
$(\sqrt r \sqrt s)^2 =  r\cdot s$
From equation 2, we have,
$r\cdot s = (p\cdot q)^2 $  ...(3)
If the equality holds the LHS of the equation must yield the same result as equation 1,
Now from equation 3,
$ \sqrt {r\cdot  s} = \sqrt{(p\cdot q)^2}$
$\sqrt{(p\cdot q)^2} = ((p\cdot q)^2)^{1/2}$ [as $\sqrt a = a^{1/2}$]
Now from the power rule $(a^b)^c = a^{b\cdot c}$, it follows that
$((p\cdot q)^2)^{1/2} = (p\cdot q)^{2/2}= p\cdot q$ .. (4)
From equation (1) and equation (4). As both sides yield the same result, it is true that, $ \forall r, s \in \mathbb{R^+}  \sqrt {r. s}  = \sqrt r \cdot \sqrt s$ $\blacksquare$
I have one concern with this proof, by this same reasoning that I have followed, I can even prove the statement for $\forall r, s \in \mathbb{R^-}$, which is not true, so I must be wrong somewhere.

Since, as you say, ''you are new to writing proofs'', I think that you want some advice to improve your proof.
The first step is to start with a good definition of the mathematical objects that you are using. In this case we need a definition of $\sqrt{a}$ for $a\in \mathbb{R}^+$. This definition is:

$1) \qquad \sqrt{a}$ is a non negative real number  $x\ge 0$ such that $x^2=a$

The second step is to well define what we want to prove, and in this case it is the identity: $\sqrt{ab}=\sqrt{a}\sqrt{b}$, where ( by the definition $1)$) we have: $\sqrt{a}=x\ge 0$ , $\sqrt{b}=y\ge 0$ and $\sqrt{ab}=z\ge 0$, such that $z^2=ab$, $y^2=b$ and $x^2=a$ (note that this exclude the possibility that $a,b$ (and $ab$) can be negative numbers).
Now, for the proof, we can start from the LHS or from the RHS of this identity. 
From the LHS:
$$
\sqrt{ab}=z \Rightarrow z^2=ab=x^2y^2
$$
So, using commutativity of the product in $\mathbb{R}$:
$$
z^2=(xy)^2
$$
and, since $xy\ge 0$ we can write
$$
xy=\sqrt{ab}=z 
$$
that is:
$$
\sqrt{a}\sqrt{b}=\sqrt{ab}
$$
Note that we don't need fractional exponents, but only the definition of the square root. in particular we did not use the rule $(a^b)^c=a^{bc}$ that is not valid, in general, for a fractional exponent.
In your proof you  started from the RHS,
I leave to you the writing of the proof in this case.
