Real Analysis sequence proof issues

I recently showed a few proofs to a professor at my school, he said that there were logical issues with my proofs and that I should stick more closely to the text book examples when constructing proofs. But he never specified exactly what aspects of the proofs needed work, and left me with more questions than I started with. Being new to proofs and Real Analysis I wanted to know which parts of my proofs are weak or wrong so I can do better in the future. Both of the proofs are on sequences.
I would like to know which parts of my proofs are not working and which parts are good. I am a self study student and don't really have a gauge on how I'm doing in my proofs. Any advice, tips, or things to change are welcomed.

Let $a_n=7-\frac{1}{\sqrt{n}}$. Show that $a_n\to7$ as $n\to \infty$.

Setup work: $|a_n-a|<\epsilon\Rightarrow\left|7-\frac{1}{\sqrt{n}}-7\right|<\epsilon\Rightarrow\left|-\frac{1}{\sqrt{n}}\right|<\epsilon\Rightarrow\frac{1}{\sqrt{n}}<\epsilon$. Solving for $n$ where $\frac{1}{\sqrt{n}}=\epsilon$ , $n=\large\frac{1}{\epsilon^2}$.

Proof. Let $\epsilon > 0$.

Set $N=1/\epsilon^2$, now for any $n > N$, $|a_n-a|<\epsilon=|7-\frac{1}{\sqrt{n}} -7|<\epsilon$ implying that $\frac{1}{\sqrt{n}}>\frac{1}{\sqrt{N}}$, now substituting $N$ into the equation $ \frac{1}{\sqrt{n}}$ produces $\large\frac{1}{\sqrt{\frac{1}{\epsilon^2}}}\Rightarrow\large\frac{1}{\frac{1}{\epsilon}}$=$\epsilon$. Thus by definition 3.7 as $n\to\infty $, $a_n \to 7$. $\mathbb{\square}$

Definition 3.7. A sequence ($a_n$) converges to $a\in \mathbb{R}$ if for all $\epsilon > 0$ there exists some $N$ such that $|a_n-a|<\epsilon$ for all $n>N$

Let ($a_n$) be a bounded sequence, and consider a second sequence ($b_n$) defined by $b_n$ := sup{$a_n,a_{n+1},a_{n+2},...$}. Prove that ($b_n$) converges.

Proof.

Since $a_n$ is defined as a bounded sequence, the supremum of $a_n$ exists, meaning $\sup\{a_n\}=U$ where $U\in \mathbb{R}$. Now for all $n \in \mathbb{N}$ the value of $\sup\{a_n\}$ will remain constant or increase, implying that $\sup\{a_n\}\leq \sup\{a_{n+1}\}$.

Since $b_n$ is defined as $b_n=\sup\{a_n,a_{n+1},a_{n+2},\dots\}$, The value of $b_n$ will also remain constant or increase, that is, $b_n \leq b_{n+1}$ implying that $b_n$ is a monotone sequence and is bounded, thus by the monotone convergence theorem $b_n$ converges for all $n$.

$\mathbb{\square}$

The first proof... It looks like you are manipulating the symbols to get $\epsilon$, but you are forgetting that you are supposed to prove what is essentially just an inequality. Namely, from $n\gt N=\frac{1}{\epsilon^2}$ you need to prove $|a_n-7|\lt\epsilon$. Now, if $n\gt N$, you can conclude:

$$|a_n-7|=\frac{1}{\sqrt{n}}\color{red}{\lt}\frac{1}{\sqrt{N}}=\epsilon$$

(the first equality you have shown correctly in your proof). I find a fault not just that your inequality sign was wrong ($\gt$ rather than $\lt$) but that you didn't seem to even care what the sign will be. Remember, the point is not to express/calculate $\epsilon$ from the given data, the point is to prove that something is/can be made smaller than $\epsilon$.

As for the second proof, it is incorrect as it incorrectly captures what is happening with those suprema. Imagine you make some set smaller: what do you expect to happen with its supremum/infimum? Answer: the supremum goes down and the infimum goes up. In our case the sets $\{a_n, a_{n+1}, a_{n+2},\ldots\}$ are getting smaller all the time: the next set is obtained by throwing away $a_n$ from the previous set. Thus, the sequence $b_n$ will be decreasing. (It will still be bounded from below, so you would still be able to apply the monotone convergence theorem.)