Prove that $\forall n \in \mathbb{Z}, \lfloor x + n \rfloor = \lfloor x \rfloor + n $.
Proof: Let $K = \{\ k\ |\ k\in\mathbb{Z},\ k \leq x+n\}$. Then, by definition,
$$ \lfloor x + n \rfloor = j = \sup(K) $$
Let $k \in K.$ Since $ n,k \in \mathbb{Z},\ k-n\in \mathbb{Z}$. In addition, $\forall k, j \geq k,$ so $j-n \geq k-n$. Therefore, if we define a new set $\ T = \{\ k-n \ | \ k -n \in \mathbb{Z},\ k \leq x+n\}= \{\ k-n \ | \ k -n \in \mathbb{Z},\ k - n\leq x\},\ j-n = \sup(T)$.
Thus, $\lfloor x \rfloor = j-n,$ or $j = \lfloor x + n \rfloor =\lfloor x \rfloor + n,$ as desired. $_{\square}$
I'm currently independently reading Apostol's Calculus for fun - while I understand the concepts, I'm not sure if I wrote an acceptable proof, or made any unjustified assumptions (I'm new at writing proofs). Even if I did prove it correctly, I feel like there's an easier way. Please point out mistakes or give any general comments/tips/suggestions to help me improve in writing proofs. Thanks!
Since you asked for tips/suggestions to improve writing proofs, I'll suggest that while I think the proof is accurate, it obscures what is going on somewhat. The problem is introducing rather intricate definitions of sets and then looking at supremums is hard to follow (for me).
Instead (since this is from Apostol), you have a previous exercise, I.3.12 #4 on page 28, in which you proved (via a method similar to the proof you give here) that for any real number $x$ there is exactly one integer $n$ which satisfies $$ n \leq x < n+1,$$ and this $n = [x]$, the greatest integer less than or equal to $x$.
Working from this definition for $[x]$ you might produce a proof for the exercise at hand that looks something like this:
Proof. Let $[x+n] = m$ for an integer $m$. Then, by the definition of the greatest integer function \begin{align*} m \leq x+n < m+1 && \implies && m-n \leq x &< m-n+1 \\ &&\implies && [x] &= m-n &(\text{again def of } [x])\\ &&\implies && [x] + n &= m. \end{align*} But, we defined $m=[x+n]$. Thus, $[x+n] = [x] + n$ for any $n \in \mathbb{Z}. \qquad \blacksquare$
I think this is easier to follow, and gives a better sense of what is happening with the greatest integer function.