Why is a square root not a linear transformation? The question says: 

Prove that the function $f(x)=\sqrt{x}$ is not a linear transformation
  (particularly $\sqrt{1+x^2}≠1+x$)

I think that this is because the exponent of $\sqrt{x}$ is $1/2$, and with that exponent, $T(cu)=c^{1/2}T(u)$, which does not follow the rules of a linear transformation...

If linearity is understood over the field of reals, then since $f(-x)$ does not exist for $x> 0$, case closed. Even if we restrict to a weak form of linearity, $\sqrt{cx} = c\sqrt{x}$ for $x> 0$ only when  $c=1$. 
Using the given hint, $(\sqrt{1+x^2})^2 - (1+x)^2 = 2x$, hence both quantities on the LHS  are equal only when $x=0$. So almost always: $(\sqrt{1+x^2})\neq (1+x)$. A single and simple counter-example using Pythagorean triples $3^2+4^2=5^2$ (from a comment): 
$$ \sqrt{1+\left( \frac{4}{3}\right)^2} = \sqrt{\left( \frac{3^2+4}{3^2}\right)} =\sqrt{\left( \frac{5^2}{3^2}\right)} = \frac{5}{3} $$
and $$1+\frac{4}{3}=\frac{7}{3}$$
and of course $$ \frac{5}{3}\neq \frac{7}{3}\,.$$
These are two reasons to conclude that $f$ is not linear. Each one is sufficient, since a counter-example suffices.
In passing, one sometimes linearize this function around $1$. A standard first order development exists when $y\sim 0$: $\sqrt{1+y} \sim 1+\frac{y}{2} $. 
