Problem. If $(X,d)$ is a metric space such that it is not complete then prove that $X$ is not compact.
My Attempt. Since $(X,d)$ is not complete, there exists a Cauchy sequence $(x_n)_{n\in\mathbb{N}}$ such that $(x_n)_{n\in\mathbb{N}}$ isn't convergent in $X$ and also such that $x_m\ne x_n$ for all $m\ne n$. Let $\mathcal{A}:=\{x_n: n\in\mathbb{N}\}$.
Claim. $\mathcal{A}$ is closed in $X$.
Proof. Otherwiese $\mathcal{A}\subset \overline{\mathcal{A}}$ (where $\overline{\mathcal{A}}$ denotes the closure of $\mathcal{A}$ in $X$). Which implies that there exists an $x\in X$ such that $x$ is a limit point of $\mathcal{A}$. Hence we conclude that there exists a subsequence $(x_{n_k})_{k\in\mathbb{N}}$ of $(x_n)_{n\in\mathbb{N}}$ such that $(x_{n_k})_{k\in\mathbb{N}}$ converges to $x$. Consequently, it follows that $(x_{n})_{n\in\mathbb{N}}$ converges to $x$, a contradiction.
If $X$ were compact, since $\mathcal{A}$ is a closed subset of $X$, it must also be compact. We will prove that $\mathcal{A}$ is not compact. Consequently it will follow that $X$ is not compact and we are done.
Observe that since $(x_{n})_{n\in\mathbb{N}}$ is not convergent, for all $x\in X$ there exists open set $U_x\ni x$ such that $\mathcal{A}\not\subseteq U_x$. In other words, for all $x\in X$ there exists open set $U_x\ni x$ such that $\mathcal{A}\cap (X\setminus U_x)\ne \emptyset$.
Now observe that, $$\mathcal{A}\subseteq X=\bigcup_{x\in X} U_x$$Since $\mathcal{A}$ is compact we have, $$\mathcal{A}\subseteq \bigcup_{i=1}^n U_{x_i}\tag{1}$$But since $\mathcal{A}\cap (X\setminus U_x)\ne \emptyset$ for all $x\in X$ we have, $\mathcal{A}\cap (X\setminus U_{x_i})\ne \emptyset$ for all $i\in\{1,2,\ldots,n\}$.
If now $A\cap\left(\displaystyle\bigcap_{i=1}^n (X\setminus U_{x_i})\right)\ne\emptyset$ then we are done because in that case there exists $a\in A$ such that $a\not\in \displaystyle\bigcup_{i=1}^n U_{x_i}$, so $(1)$ cannot follow.
But I can't resolve the case when $A\cap\left(\displaystyle\bigcap_{i=1}^n (X\setminus U_{x_i})\right)=\emptyset$. Can anyone help?
You can't just choose any such sets $U_x$; you have to choose them carefully such that no finite subcollection of them can cover $\mathcal{A}$. (Also, you don't need to choose $U_x$ for every $x\in X$, only for every $x\in\mathcal{A}$.) In fact, you can choose $U_x$ for each $x\in\mathcal{A}$ such that $U_x\cap\mathcal{A}=\{x\}$ for each $x$, and then it is clear that no finite subcollection can cover $\mathcal{A}$.
To find such a $U_x$, just note that you can take $U_x=X\setminus(\mathcal{A}\setminus\{x\})$. To prove that this is open, note that your argument that shows that $\mathcal{A}$ is closed in $X$ actually also applies to any subset of $\mathcal{A}$, so in particular $\mathcal{A}\setminus\{x\}$ is closed in $X$.