This might seem like a duplicate, but I couldn't find anyone proving it like this. I feel like my proof is wrong because I couldn't find anyone doing it this way. It would be helpful if anyone can verify it for me.
Let's lay down some definitions. Let $X$ be a topological space and $A \subset X$.
By $\bar{A}$, I mean the closure of $A$. By $\text{int}(A)$, I mean the interior of $A$.
And here's probably the most important definition in this question.
By $\partial A$, I mean the boundary of $A$ and I choose the definition that $\partial A = \bar{A} \backslash \text{int}(A)$.
Using this definition, I found that it was rather easy to prove the following claim.
Claim: Let $X$ be a topological space and $A \subset X$. Then $\bar{A} = \text{int}(A) \cup \partial A$
Proof. By definition of the boundary, we have that \begin{equation} \text{int}(A) \cup \partial A = \text{int}(A) \cup (\bar{A} \backslash \text{int}(A)) \end{equation} Observe that $\bar{A} \backslash \text{int}(A) = \bar{A} \cap (X \backslash \text{int}(A))$. Moreover, it is true for any sets $S, T, U$ that $S \cup (T \cap U) = (S \cup T) \cap (S \cup U)$. Thus,
\begin{equation} \text{int}(A) \cup (\bar{A} \backslash \text{int}(A)) = \text{int}(A) \cup \{ \bar{A} \cap (X \backslash \text{int}(A))\} = \{ \text{int}(A) \cup \bar{A} \} \cap \{ \text{int}(A) \cup (X \backslash \text{int}(A))\} \end{equation} Now, $\text{int}(A) \cup (X \backslash \text{int}(A)) = X$. Also, we have that $\text{int}(A) \subset A \subset \bar{A}$. Thus, we have that \begin{equation} \text{int}(A) \cup (\bar{A} \backslash \text{int}(A)) = \text{int}(A) \cup \bar{A} = \bar{A} \end{equation} But the LHS is just $\text{int}(A) \cup \partial A$, so we have really proven \begin{equation} \bar{A} = \text{int}(A) \cup \partial A \end{equation} as required.
By definition of closure and interior, $$ A\subset \bar{A},\quad \text{int}(A)\subset A $$ then $\text{int}(A)\subset \bar{A} $.
Now $\partial A = \bar{A} \backslash \text{int}(A)$ implies $\bar{A} = \text{int}(A)\cup \partial A $ simply by an identity on set operations:
$X=Y\setminus Z\quad \textrm{and} \quad Z\subset Y \Rightarrow Y=X\cup Z$