Show that $\lbrace x_n : n \in \mathbb{N} \rbrace \cup \lbrace x \rbrace$ is a compact subset of $(X,d)$ 
Let $(X,d)$ be a metric space, and $(x_n)_{n \in \mathbb{N}}$ a
  sequence in $X$, with $x_n \rightarrow x \in X$ (w.r.t the usual
  metric $d$). Show that $\lbrace x_n : n \in \mathbb{N} \rbrace \cup
\lbrace x \rbrace$ is a compact subset of $(X,d)$

Need step by step proof!
These are my thoughts:
I know to show $(X,d)$ is a compact subspace I have to show that every open cover of $X$ has a finite subcover.  
This means I must first show $(X,d)$ is an open cover of (what?)
But, isn't the singleton $\lbrace x \rbrace$ closed in $(X,d)$? Also, I suspect this a gross misunderstanding, but if $x_n \rightarrow x \in X$ (w.r.t the usual metric $d$), then is it wrong to interpret the union in question to be:
$\lbrace x \rbrace \cup \lbrace x \rbrace$?  
 A: For an open cover of $\{x_n\mid n\in\mathbb{N}\}\cup\{x\}$ there is an open set $U$ from the cover with $x\in U$ and because $x_n\to x$ there is a $N\in \mathbb{N}$ with $x_m\in U$ forall $m>N$. So when we chose for every $m\leq N$ an open set $U_m$ from the cover with $x_m\in U_m$  we get a finite subcover by the open sets $U_m, m\leq N$ and $U$ and so $\{x_n\mid n\in\mathbb{N}\}\cup\{x\}$ is compact.
A: Hint: Let $S:=\{x_n: n\in \mathbb N\}\cup \{x\}$. Since $(X,d)$ is a metric space, then showing that $S$ is compact is equivalent to showing that $S$ is sequentially compact: that every sequence in $S$ has a convergent subsequence whose limit is in $S$. Now the convergence of $x_n$ to $x$ implies that any subsequence of $x_n$ converges also to $x$. 
A: If $\lim_{n\to \infty}d(x,x_n)=0,$ then, by the def'n of $\lim$, whenever $r>0$, the set $\{n: x_n\not \in B_d(x,r)\}$ is finite (where $B_d(x,r)=\{y: d(x,y)<r\}$).
$U$ is an open subset of $X$ iff for every $y\in U$ there exists $r_y>0$ such that $B_d(y,r_y)\subset U.$
For brevity let $A=\{x\}\cup \{x_n: n\in N\}.$ Let $C$ be an open cover of $A.$ There exists $U_0\in C$ with $x\in U_0,$ and there exists $r_x>0$ such that $B_d(x,r_x)\subset U_0$.
Let $S=\{n:x_n\not \in B_d(x,r_x)\}.$ Then $S$ is finite. For each $n\in S$ there exists $U_n\in C$ such that $x_n\in U_n.$
Then $C^*=\{U_0\}\cup \{U_n:n\in S\}$ is a finite subset of $C.$ And $C^*$ is a cover of $A .$ Because: (i). $x\in U_0.$ (ii) If $n\not\in S$ then $x_n\in B_d(x,r_x)\subset U_0.$ (iii). If $n\in S$ then $x_n\in U_n.$ 
Remark. It is useful that, in a metric space, a sequence $(x_n)_{n\in N}$ converges to $x$ iff $\{n:x_n\not \in U\}$ is finite for every nbhd $U$ of $x.$
