Suppose that $(X, \tau_1)$ is compact and $\tau_1 \subset \tau_2$. Is the space $(X, \tau_2)$ compact? 
Suppose that $(X, \tau_1)$ is compact and $\tau_1 \subset \tau_2$. Is the space $(X, \tau_2)$ compact? Does the converse hold i.e if $(X, \tau_1)$ is compact and $\tau_2 \subset \tau_1$?

The first one shouldn't hold since if $X= [0,1]$ and $\tau_1$ is the usual topology of $\Bbb R$, then I think that if $\tau_2$ is the lower limit topology we have $\tau_1 \subset \tau_2$ and $X$ wouldn' t be compact?
The second one also doesn't seem true. If $(X, \tau_1)$ is compact and $\tau_2 \subset \tau_1$, then every open cover of $X$ has a finite subcover, but I don't think why this would hold for the coarser topology $\tau_2$? I think it could be that $\tau_2$ doesn't have "enough" elements to satisfy this.
 A: I think your confusion for the second one arises from the meaning of "finite subcover". A finite subcover is a finite cover consisting of elements of the initial cover. So, if you cover $X$ by elements of $\tau_2$, these elements are also elements of $\tau_1$ and hence give a finite subcover, which is what you need.
A: The first one holds if $X$ is finite, since every topology on finite sets are compact. If $X$ is infinite, for any topology $\tau_1$ on $X$ one can take $\tau_2$ to be the discrete topology, which is non-compact.
On the other hand, if $(X, \tau_1)$ is compact and $\tau_2\subset \tau_1$, then $(X, \tau_2)$ is also compact: take any covering $\mathscr U \subset \tau_2$ of $X$. Then it is also a covering of $(X, \tau_1)$ and hence has a $\mathscr U_1 \subset \mathscr U$.
A: Suppose, $card(X) \ge \aleph_{0}$
Consider two topology on $X$
$\tau_{indiscrete}=\{\emptyset, X\}$ and
$\tau_{discrete}={\scr{P}}(X) $
Then, $\tau_{indiscrete}\subset\tau_{discrete}$

Claim :$(X, \tau_{indiscrete}) $ is compact but $(X,\tau_{discrete}) $
is not compact.

$X\subset (X,\tau_{indiscrete}) $
Suppose, $X\subset \cup_{\alpha} U_{\alpha}$ , then $U_{\alpha} =X$ for at least one $\alpha$.
$X\subset (X, \tau_{indiscrete}) $
Then, $X\subset \cup_{x\in X} \{x\}$
has no finite subcover otherwise it would contradict cardinality of $X$.
For your second question,
If $X$ is compact in a topology then it is also compact in any of it's coarser topology.
$\tau_{1}\subset \tau_{2}$
$(X, \tau_{2}) $ is compact.
Suppose, $X\subset \cup_{\alpha} U_{\alpha}$
Where, $U_{\alpha}\in \tau_{1} \subset \tau_{2}$
Then there exists a finite subcover i.e a $I$ finite index set such that
$X\subset\cup_{i\in I}U_{\alpha_{i}}$
Where, $U_{\alpha_{i}}\in \tau_{2}$
Then, $\{U_{\alpha_{i}}:i\in I\}$ is a finite subcover of $\{U_{\alpha}\}$.
Hence, $(X, \tau_{1}) $ is also compact.
A: $X$ being $\tau_1$-compact means that for every net there exists a $\tau_1$-convergent subnet. If $\tau_2 \subseteq \tau_1$, convergence in $\tau_1$ implies convergence in $\tau_2$. Therefore, for every net there also exists a $\tau_2$-convergent subnet, and thus $X$ is $\tau_2$ compact. The converse is not true.
