Let $f\colon X \rightarrow Y$ be a continuous and surjective function. Show that if $X$ is compact, then so is $Y$. Here is my attempt at answering the above question. 
(I feel that there are gaps in my knowledge in this topic and don't have a sound understanding of what a covering actually is but here goes!) 
As f is surjective, there are $x_1, x_2∈ X$ s.t $f(x_1)=y_1, f(x_2)=y_2$. And f is continuous if for every open $v⊆Y, f^{-1}(v)$ is open in X.
If X is compact, then every open covering $U=(u_i)_{i∈I}$ has a finite subcovering. 
Therefore, $f^{-1}(y_1)=x_1$ which is also an open covering. Thus Y is compact. 

Correct me if wrong:
Hopefully adding a bit of detail to José Carlos' solution .
1) Let $O_i$, $i \in I$, be an open cover of $Y$, 
i.e. $(\cup O_i)_{i \in I} =Y.$
$X \subset f^{-1}Y$ , or $X \subset f^{-1}(\cup O_i)_{i \in I}$.
2)$X \subset f^{-1}(\cup O_i)_{ i \in I}=$
$(\cup f^{-1}(O_i))_{i \in I}$.
3) Since $f$ is continuous, 
$f^{-1}(O_i)_{ i \in I}$ is open, and
$\cup f^{-1}(O_i)_{i \in I}$ is an open cover of $X$
4) Since $X$ compact there is a finite subset $I_f$ of $I$, such that
$X \subset \cup f^{-1}(O_i)_{i \in I_f}$.
5) $Y=f(X)=f(\cup f^{-1}(O_i)_{ i \in I_f})=$
$\cup f(f^{-1}(O_i)_{i \in I_f})= \cup (O_i)_{i \in I_f}$, i.e
a finite cover of $Y$.
Used $f(f^{-1})(O_i)=O_i$ , $i \in I_f$, since $f$ surjective.
