Prove that $T$ is a tree of order $n$ if and only if $T$ is a connected graph with size $n-1$.
Here is my answer.
Proof: ($\Rightarrow$) Let $T$ be a tree of order $n$ and let the size of $T$ be $m$. By definition, $T$ is connected. Also $n = m + 1\implies m = n - 1$, so the tree has size $n - 1$.
($\Leftarrow$) Let $T$ be a connected graph with size $n-1$. It suffices to show that $T$ is acyclic or has no cycles. If $T$ contains a cycle $c$, and $e$ is an edge of $c$, then $T-e$ is a connected graph of order $n$ having size $n-2$, which is impossible. Therefore $T$ is acyclic or has no cycles. Thus, $T$ is a tree.
I am not sure my answer is good so I am just wondering if I can get some feedback if it is correct.
It seems to me like you are assuming the result you want to prove in both cases. For $\Rightarrow$, how do you know $n=m+1$? For $\Leftarrow$, why is it impossible for a connected graph of order $n$ to have $n-2$ edges?