How does this prove: All eigenvalues of a triangular matrix = All its diagonal entries? Source: Linear Algebra by David Lay (4 edn 2011). p. 269 Theorem 5.1.1.

For simplicity, consider the $3\times 3$ case. If $A$ is upper triangular, then 
  $ A-\lambda I=
\begin{bmatrix}
a_{11} & a_{12} & a_{13}\\
0 & a_{22} & a_{23}\\
0 & 0 & a_{33}
\end{bmatrix} 
- 
\begin{bmatrix}
\lambda & 0 & 0\\
0 & \lambda & 0\\
0 & 0 & \lambda
\end{bmatrix}
=\begin{bmatrix}
a_{11}-\lambda & a_{12} & a_{13}\\
0 & a_{22}-\lambda & a_{23}\\
0 & 0 & a_{33}-\lambda \\
\end{bmatrix}$
$\lambda$ is an eigenvalue of $A \iff$ The equation $(A-\lambda I)x=0$ has a
  nontrivial solution. $\iff$ $(A-\lambda I)x=0 $ has a free variable $ \iff $
  Because of the zero entries in $A-\lambda I$,  at least one of the entries on the diagonal of $A-\lambda I$ is zero.
  $\color{red}{\iff} \lambda$ equals $\color{red}{one \, of } $ the entries $a_{11},\ a_{22},\ a_{33}$ in $A$. 
  For the  lower triangular case, see Question 5.1.28.

$1.$ I don't understand the red $\color{red}{\iff}$. $a_{11} - \lambda = 0 \iff$ The first column is $\mathbf{0}. \iff$ $x_1$ is a free variable. But what about the other entries? 
$2.$ I linked an analogous question. How does the proof overhead proves that all of the eigenvalues = all its diagonal entries, when it states $\color{red}{one \, of } $? 
 A: It's equivalent to showing that an upper triangular matrix is injective iff none of its diagonal entries are $0$.
If one of the diagonal entries, say the $i$th, is $0$, then restricting the operator to the span of the first $i$ basis vectors gives a map into the span of the first $i-1$ basis vectors, which by Rank-Nullity cannot be injective.
Now, if all of the diagonal entries are nonzero, then the columns must be independent: the last column is the only one with a nonzero $n$th entry, so its coefficient has to be $0$. Continuing this way, you can kill all the coefficients.
A: Not sure I get your question perfectly but :
$\lambda$ is an eigenvalue of A $\Leftrightarrow  \lambda $ is a root of the characteristic polynom of A $\Leftrightarrow  |A -\lambda Id| =0 \Leftrightarrow \Pi_{i}^n (a_{ii}-\lambda)=0 \Leftrightarrow \exists i\  \lambda =a_{ii}$. 
Most equivalence are quite obvious ( you need to know that the determinant of a tringular matrix is the  product of its diagonal elements and that the kernel of a matrix which   determinant is 0  has a non trivial kernel). 
A: Expand $\det(A - \lambda I)$ using minors of the first column.
The only term that survives is $(a_{11}-\lambda)\det(B)$ where:
$B = \begin{bmatrix}a_{22}-\lambda&a_{23}\\0&a_{33}-\lambda \end{bmatrix}$, which clearly has determinant:
$(a_{22} - \lambda)(a_{33} - \lambda)$.
A similar strategy works for any $n \times n$ upper triangular matrix.
This shows that every eigenvalue (root of $\det(A - \lambda I)$) is a diagonal entry of $A$ and vice-versa.
Surely you can see that (in the $3\times3$ case) if $a_{33} - \lambda = 0$ that the last ROW is $0$, recall column rank = row rank.
If $a_{22} - \lambda = 0$, then the 2nd row is a scalar multiple of the 3rd row, so after row-reduction, we'll have at LEAST one zero row at the bottom.
The logic is a bit more involved for an $n \times n$ upper triangular matrix, but if one of the diagonal elements of $A - \lambda I$ is $0$, it should be clear that THAT row is a linear combination of the rows below it.
A: Let $\lambda$ be a diagonal entry of $A.$ In $A-\lambda I,$ the corresponding diagonal entry is $0.$ Look at the first $0$ entry on the diagonal of $A-\lambda  I.$ (first from $(1,1)$th entry through $(n,n)$th.) If this occurs on the $k$th column, then the $k$th column is a linear combination of earlier columns. Thus ${\rm rank}(A-\lambda I)<n.$ And, ${\rm null}(A-\lambda I)\geq 1.$ Thus $Av=\lambda v$ for some $v\neq 0.$ 
