Are these implications regarding compactness correct?

Question

Is the following true and if not how is it incorrect :

a subset of a metric space being compact $\Leftarrow\Rightarrow$ it is complete and totally bounded $\Rightarrow$ it is closed and totally bounded.

Answer 1

Your statement is essentially correct, but can be stated more precisely:

Let $(X,d)$ be a metric space, and let $A \subset X$ be a metric subspace, so we give $A$ the metric $d_A$ induced from $d$ (just the restriction of $d$ to $A \times A$).

Then indeed $(A, d_A)$ is compact iff it is complete and totally bounded. The compactness (or completeness plus total boundedness) does imply that $A$ is closed in $(X,d)$ (and also still totally bounded), but $A$ being closed in $(X,d)$ and totally bounded (under $d_A$) does not imply compactness of $(A,d_A)$. This would be true if $(X,d)$ were complete to start with, but not otherwise, e.g. take $X=(0,1)$ in the Euclidean metric and $A=X$, which is closed in $X$ (trivially) and totally bounded, but not compact.