For some reason I have a hard time proving the following lemma
Let $\varphi:G \to GL(V)$ be a representation.
dim Hom$_{G}(\varphi, \varphi) =1 \implies \varphi$ is irreducible
I thought about using schur's lemma, but that wont'work, since it holds for the representation already being irreducible. Moreover, I know that if dim Hom$_{G}(\varphi, \varphi) =1 $ hold, it would mean that $T\varphi_{g} = \varphi_{g} T$ with $T \in$ Hom$_{G}(\varphi, \varphi)$, would imply that $T$ has the form of $\lambda I$. And since the kernel is trivial, and the image is $V$, I would not have to worry about those, but how do I prove that those are in fact the only $G$-invariant subspaces of $V$ under $\varphi$ ?
This is not in general true. Let $G=\left\{\begin{pmatrix}1&a\\0&b\end{pmatrix}:a,b\in\mathbb{C},b\neq 0\right\}\subset GL(\mathbb{C}^2)$, and let $V=\mathbb{C}^2$. This is not irreducible because $W=\mathbb{C}\begin{pmatrix}1\\0\end{pmatrix}$ is a non-trivial subrepresentation.
Suppose that $f:V\to V$ commutes with the $G$ action. Since $$\begin{pmatrix}1&1\\0&1\end{pmatrix} f \begin{pmatrix}1\\0\end{pmatrix} = f\begin{pmatrix}1&1\\0&1\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix} = f \begin{pmatrix}1\\0\end{pmatrix},$$ we see that $f\begin{pmatrix}1\\0\end{pmatrix}$ is an eigenvector of this Jordan block, hence $f\begin{pmatrix}1\\0\end{pmatrix}=\lambda \begin{pmatrix}1\\0\end{pmatrix}$ for some $\lambda\in\mathbb{C}$. Since $$\begin{pmatrix}1&0\\0&b\end{pmatrix} f \begin{pmatrix}0\\1\end{pmatrix} = f\begin{pmatrix}1&0\\0&b\end{pmatrix}\begin{pmatrix}0\\1\end{pmatrix} = bf \begin{pmatrix}0\\1\end{pmatrix},$$ by again considering eigenvectors, $f\begin{pmatrix}0\\1\end{pmatrix}=\mu \begin{pmatrix}0\\1\end{pmatrix}$ for some $\mu\in\mathbb{C}$. Hence $$f=\begin{pmatrix}\lambda&0\\0&\mu\end{pmatrix}.$$ But, consider the fact that $$\begin{pmatrix}1&1\\0&1\end{pmatrix} \begin{pmatrix}\lambda&0\\0&\mu\end{pmatrix} - \begin{pmatrix}\lambda&0\\0&\mu\end{pmatrix} \begin{pmatrix}1&1\\0&1\end{pmatrix} = \begin{pmatrix}0&\mu-\lambda\\0&0\end{pmatrix}.$$ Hence, $\mu=\lambda$ since by hypothesis $f$ commutes with the $G$ action. Therefore $f=\lambda\;\mathrm{id}_V$.
So we have a finite-dimensional $G$ representation whose endomorphism ring is $\mathbb{C}$ yet is not irreducible.
Of course, if $G$ is a finite group, then if there were a subrepresentation $W\subset V$, we can obtain a complementary subrepresentation $W'\subset V$ such that $V=W\oplus W'$, since one can average over the group to produce a projection that commutes with the $G$ action. The operators $f:W\oplus W'\to W\oplus W'$ given by $f(w,w')=(\lambda w,\mu w')$ for $\lambda,\mu\in\mathbb{C}$ commute with the $G$ action. If both $W$ and $W'$ are nontrivial, this would mean $\mathbb{C}^2\subset\hom_G(\phi,\phi)$. But if $\hom_G(\phi,\phi)$ is $1$-dimensional, then at least one of $W$ or $W'$ is $0$-dimensional. It would follow that $V$ is irreducible.