I have a transformation $T$ such that $T(f)=g$, where $g= tf'(t)$
And, I have to prove why every real $\lambda$ is eigenvalue for $T $
First of all ,
I set $λf(t)=t f'(t)$
and I found out that $f(t)=Ct^λ$ , where $C$ is nonzero number.
It seems so obvious, by just looking at $f(t)$, that $\lambda$ is every real number.
But when It comes to actual proof of why $\lambda$ is every real, I am stuck...
You have most of the right ideas, but you've scrambled them around so that it no longer looks like you answered the question.
We have a linear transformation $T$ over differentiable functions from $\Bbb R$ to $\Bbb R$ defined by $$ T[f(t)] = tf'(t) $$ We say that a value $\lambda \in \Bbb R$ is an eigenvalue of $T$ if there exists a non-zero vector (i.e. function) $f:\Bbb R \to \Bbb R$ such that $T[f(t)] = \lambda \,f(t)$. That is, $\lambda$ is an eigenvalue if there exists a non-zero solution to the equation $$ tf'(t) = \lambda f(t) \implies\\ \frac{f'(t)}{f(t)} = \frac{\lambda}{t} $$ We can solve this by integrating both sides to find the solution $$ \log(f(t)) = \lambda \log(t) + C \implies\\ f(t) = C t^\lambda $$ So, for a value $\lambda$, the function $f(t) = t^\lambda$ is an eigenvector associated with $\lambda$, making $\lambda$ an eigenvalue of $T$.
Note that the statement "every real number is an eigenvalue" (which looks like the statement you need to answer) is not the same as the statement "every eigenvalue is real".