So, i am going to show that, by using $$(T5): |E|=|V|-1$$ that the tree T has at least two vertices with degree 1.
My attempts so far:
We know that the sum of all degrees of a tree with n vertices over all v is
$$ \sum\limits_{v}deg(v)=2|E|=2(n-1)$$ Furthermore, we know that the number of vertices is $$ \sum\limits_{i=1}V_i=n$$ there $V_i$ is the number of vertices with degree i. This means that $deg(v)=iV_i$ and by that we can rewrite the first sum as $$ \sum\limits_{i=1}iV_i=2|E|=2(n-1)$$ Now, if we subtract the second sum from the sum above, we will get $$ \sum\limits_{i=1}(i-1)V_i = 2(n-1)-n = n-2$$
Here it takes stop for me. And i am not clearly sure why i should subtract the second sum from the first one. The first one is the sum of all degrees over vertices v and the second one is the sum of all vertices $V_i$ with the degree i,namely the number of vertices in the tree. But what does this subtraction gives me? and how can i show that at least 2 of the verticec has degree 1?
The effect of the subtraction is that $V_1$ does not contribute to the sum (because $i-1=0$ for $i=1$), whereas all $V_i$ with $i>1$ contibute at least with a factor of one. In fact, I would subtract twice and in the other direction: $$ V_1\stackrel{(1)}\ge \sum_{i\ge1} (2-i)V_i=2n-2(n-1)n=2$$ where $(1)$ holds because on the right hand side, $V_1$ occurs with a factor of $2-1=1$ as on the left, and all other (nonnegative!) $V_i$ are added with a factor $2-i\le 0$, so cannot cause an increase.
Remark: What about vonbrands comment? Well, the above implicitly assumes that $V_0=0$. Strictly speaking we should start the summation at $i=0$, which then only gives us $$ 2V_0+V_1\ge \sum_{i\ge0} (2-i)V_i=2n-2(n-1)n=2$$ i.e.: $V_0\ge 1$ or $V_1\ge 2$.