Is the closure of a set always complete?

Question

Suppose we have a metric space $(X,d)$ that is not complete. (i.e. there are Cauchy sequences that do not converge). Would the metric space $(\bar{X}, d)$ (where $\bar{X}$ denotes the closure of $X$) be a complete metric space?

My thought process is that the closure of a set contains all its limit points but this is not convincing me as much as I'd like.

Is what I am suggesting true and also is there a formal proof for it?

EDIT: As others have pointed out, $\bar{X}$ makes no sense in the way I phrased it. So as @Randall mentioned I will rephrase the question to:

Let $(X,d)$ be a metric space and suppose $A⊆X$. Give A the induced metric and suppose A is not complete. Then is it true that $\bar{A}$ is complete with the induced metric?

I know that the closure will be complete if $(X,d)$ is complete in the first place but without saying anything about $X$ can I say the closure of a subset of $X$ is complete?

For example I think interval $(0,1)$ is not complete in $\Bbb{R}$ with the standard Euclidian distance metric, but its closure, $[0,1]$, could be complete.

Answer 1

No, it is not true. If it was true, then every metric space $(X,d)$ would be complete, since $\overline X=X$. But there are non-complete metric spaces.

Answer 2

The question is bad posed. Any topological space is both open and closed in itself, thus if $X$ is a metric space which is not complete, then his closure (which is $X$ itself) is not complete.

Given a subspace $A$ of $X$ endowed with the same metric, then there is no need for its closure to be complete. Bare in mind that limit points means every neighborhood interesects both $A$ and its complementary, it is nothing related to the existence of the limit for all cauchy sequences.

As for your last question, yes $(0,1)$ is not complete whereas $[0,1]$ is complete due to Heine-Cantor theorem.