Let $\mathbb{K}$ be a field, $1\leq n\in \mathbb{N}$ and let $V$ be a $\mathbb{K}$-vector space with $\dim_{\mathbb{R}}V=n$. Let $\phi :V\rightarrow V$ be a linear map.
The following two statements are equivalent:
There is a basis $B$ of $V$ such that $M_B(\phi)$ is an upper triangular matrix.
There are subspaces $U_1, \ldots , U_n\leq_{\mathbb{K}}V$ such that $U_i\subset U_{i+1}$ and $U_i$ is $\phi$-invariant.
Let $\phi$ satisfy the above properties. Then show that there is $0\neq v\in V$ and a $\lambda\in \mathbb{K}$ such that $\phi (v)=\lambda v$.
For that I have done the following:
We consider the subspace $U_1$. Since $U_1$ is $\phi$-invariant, it follows for $v\in U_1\subset V$ that $\phi (v)=\lambda v$, with $\lambda\in \mathbb{K}$.
Is that correct and complete?
You statement is correct if you assume that dim $U_{1} = 1$, which is not specified.
If $\phi$ satisfies one of the two, thanks to the equivalence in particular satisfies the first. Take the basis $\mathcal{B} = \left\lbrace v_{1},\cdots,v_{n}\right\rbrace$ in which $\mathcal{M}_{\mathcal{B} \to \mathcal{B}}(\phi)$ is upper triangular, and take $v_{1}$. Do you see what $\phi(v_{1})$ is equal to ?