Let $\psi:G \longrightarrow GL(V)$ be equivalent to a completely reducible representation. Then $\psi$ is completely reducible.
Idea: Since $\psi$ is equivalent to a completely reducible representation. Suppose that $\phi:G \longrightarrow GL(W)$ is such that $\phi\cong\psi$ and $T:V\longrightarrow W$ a vector space isomorphism with $\psi_g=T^{-1}\phi_gT$, $ \ $ $\phi\cong\phi^{(1)}\oplus\phi^{(2)}\oplus...\oplus\phi^{(n)}$ where $\phi^{(i)}$ are irreducible representations. Then I need show that $\psi\cong\psi^{(1)}\oplus\psi^{(2)}\oplus...\oplus\psi^{(n)}$ where $\psi^{(i)}$ are irreducible representations.
$\oplus$ (direct sum).
$\cong$ (equivalent).
I am new to mathematics, and I still do not have enough intuition in the theory of group representations. I don't know how to continue.
Can you help me please? any suggestions are welcome.
The single biggest mistake I see inexperienced math students make is ignoring the definitions. In particular, you haven't specified what a direct sum of representations means. If you did, you would assume that $\psi=\psi^{(1)}\oplus\psi^{(2)}$ (not necessarily irreducible) and apply the definition to show that equivalence implies that $\phi=\phi^{(1)}\oplus\phi^{(2)}$. By induction, you can deduce that if $\psi$ is a direct sum of irreducible representations, then so is $\phi$.
To indicate how such a proof would go, I'll give my own (equivalent) definition of a completely reducible representation and use it as the basis of a proof. As an exercise, I encourage you to apply the definitions you've received in class to get the result on your own.
Given a representation $\psi:G\to GL(W)$, call a subspace $X\subset W$ $\psi$-invariant if $\psi_g(X)\subset X$ for all $g\in G$. Then, we say that $\psi:G\to GL(W)$ is completely reducible if, whenever $X\subset W$ is $\psi$-invariant, there exists $Y\subset W$ such that $Y$ is $\psi$-invariant and $W=X\oplus Y$.
Now, assume that $\phi:G\to GL(V)$ is equivalent to $\psi:G\to GL(W)$, so that $\psi_g=T^{-1}\phi_gT$ for some isomorphism $T:W\to V$. Assume that $\psi$ is completely reducible.
To show that $\phi$ is completely reducible, appeal to the definition. Assume that $X\subset V$ is $\phi$-invariant. We will use $T$ to construct $Y\subset V$ satisfying the definition. To this end, let $X'=T^{-1}(X)\subset W$. Then, $X'$ is $\psi$-invariant since, for all $g\in G$, $\phi_g(X)\subset X$ and, therefore, $$ \psi_g(X')=\psi_g(T^{-1}(X))=T^{-1}(\phi_g(TT^{-1}(X)))=T^{-1}(\phi_g(X))\subset T^{-1}(X)=X'.$$ By assumption, $W$ is $\psi$-invariant. By the definition, there exists $Y'\subset W$ with $Y'$ $\psi$-invariant and $W=X'\oplus Y'$. Let $Y=T(Y')\subset V$. Then, $Y'$ is $\phi$-invariant since $$ \phi_g(Y)=\phi_g(T(Y'))=T(T^{-1}(\phi_g(T(Y')))=T(\psi_g(Y'))\subset T(Y')=Y. $$ Finally, we have to show that $V=X\oplus Y$. By definition, this means that $X\cap Y=0$ and $X+Y=V$. Well, $$X\cap Y=T(T^{-1}(X\cap Y))\subset T(T^{-1}(X)\cap T^{-1}(Y))=T(0)=0.$$ Finally, if $v\in V$, then $T^{-1}(v)\in W$, so $T^{-1}(v)=x'+y'$ for some $x'\in X'$ and $y'\in Y'$ (since $W=X'\oplus Y'$). Set $x=T(x')$ and $y=T(y')$. Then $$x+y=T(x')+T(y')=T(x'+y')=T(T^{-1}(v))=v.$$ Hence, $V=X+Y$ as required.