A function $g(x)$ has one and only one real root if $g'(x)\leq k <0$. $g : \mathbb{R} \to \mathbb{R}$ is differentiable on $\mathbb{R}$.  Then   $g(x)$ has one and only one real root if $g'(x)\leq k <0$. 
Proof attempt:
Let us assume the contrary, i.e. $g(x)$ has no real zero at all. 
Therefore, being continuous, $g(x)$ cannot be both positive and negative on $\mathbb{R}$ . So, firstly we assume that $g(x) <0$ for every $x \in \mathbb{R}$. 
We take some $a>0$. (WLOG, take $a=1$). Now, $g(1)/1<0$ and $g(1)/k >0$. So, $\displaystyle\frac{g(-g(1)/k)-g(1)}{-g(1)/k-1} \leq k <0 \implies -k \leq \displaystyle\frac{g(-g(1)/k)-g(1)}{g(1)/k+1} \implies 0<-k \leq g(-g(1)/k) $ 
So, we have found at least one  point in the domain, where $g(x)$ is positive. So, $g(x)$ must have a zero. Now, $g'(x)<0, \ \  \forall x \ \in \mathbb{R}$ makes the function one-to-one. 
[Note that the numerator must be positive, since $1>- g(1)/k \implies g(-g(1)/k)>g(1)$]
For the assumption that $g(x)>0$, we consider the points $a<0$ and $-g(a)/k$ (WLOG, take $a=-1$). Everything else is kept the same. 
Are the statement and the proof both correct, or is there any mistake? 
Please verify. 

We can prove that: 
$$\lim_{x \to -\infty} g(x) = \infty \quad (1)$$
$$\lim_{x \to \infty} g(x) = -\infty \quad (2)$$
Lets prove the second statement, the first statement can be proved using a similar argument. 
Let's assume the opposite, that: 
$$\lim_{x \to \infty} g(x) = C \neq -\infty$$
Now if we consider the following limit: 
$$\lim_{x \to \infty} g'(x) = \lim_{x \to \infty}\left(\lim_{h \to 0} \frac{g(x+h)-g(x)}{h}\right)$$
$$=\lim_{h \to 0}\left(\lim_{x \to \infty} \frac{g(x+h)-g(x)}{h}\right)$$
$$=\lim_{h \to 0}\frac{C-C}{h}$$
$$=0$$ 
Which contradicts the fact that $g'(x) < 0$ for all $x$. 
Since $g'(x) < 0$, $g(x)$ can't go to $+\infty$ as $x \to \infty$, so it goes to $-\infty$.
This proves that $g$ has only one real root. 
