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...

I suppose you could start with the definition of a linear transformation.

I think you have it right that $T(cu)\ne c T(u)$ for scalar $c$.

You might look at the other property of a linear transformation:

Is $T(x+y) = T(x)+T(y)$?

(i.e. does $\sqrt{x+y}=\sqrt{x}+\sqrt{y}$?)

Edit: If we want to be more abstract. (I doubt this is what the question is aiming for, but others have pointed this out.) We can define addition in our vector space as multiplication and scalar multiplication as exponentiation. This would work for $\mathbb{R}_{>0}$. We can then show that the $f$ does work. It is likely the problem you are working on intends the reals in which case you can ignore everything past "edit".

Others have gone the symbolic route... However, it is sufficient to show one (valid) assignment of the variables that fails to achieve equality. For instance, if $x = 1$, we would need to show that $\sqrt{1+1^2} = \sqrt{2}$ does not equal $\sqrt{1} + \sqrt{1^2} = 1 + 1 = 2$. And this should be evident.