Prove that a set consisting of a sequence and its limit point is closed

As I said in my comment, your proof is ok. But here is the way I would have done it. I'll write the proof in a more formal way, because in math you can only talk loose after you master writing properly.

Let $y\in X\setminus A$. Let $\varepsilon=d(y,x)/2$. Then, by the convergence $x_n\to x$, there exists $n_0$ such that $x_n\in B_\varepsilon(x)$ for all $n\geq n_0$. So, for $n\geq n_0$, $$\tag{1} d(x_n,y)> d(x,y)-d(x_n,x)>d(x,y)-\varepsilon=d(x,y)/2. $$ Let $\delta=\min\{d(x,y),d(x_1,y),\ldots,d(x_{n_0},y)\}/2$. Then $d(y,x_n)\geq\delta$ if $n\leq n_0$, and by ($1$) $d(y,x_n)\geq\delta$ if $n\geq n_0$. This shows that $B_\delta(y)$ has no intersection with $A$, i.e. $B_\delta(y)$ is contained in $X\setminus A$. As $y$ was arbitrary, this shows that $X\setminus A$ is open, i.e. $A$ is closed.