Homeomorphism preserve compactness and conectedness I would like some feedback in my proof for the following question:

Let $M$ and $M{'}$ be homeomorphic metric spaces. Show the following:
      (1) If $M$ is compact, then $M{'}$ is compact.
      (2) If $M$ is connected, then $M{'}$ is connected. 

My atempt: 
(1) Let $\mathbb{G}=\{G_{\lambda}:\lambda \in \Lambda\}$ be an open cover of $M{'}$. Since $M$ and $M{'}$ are homeomorphic, the function $f:M \to M{'}$ is continuous, bijective and has continuous inverse.
By the continuity of $f$, $f^{-1}\big(\bigcup_{\lambda \in \Lambda}G_{\lambda}\big)$ is an open cover of $M$. 
Since $M$ is compact, $\exists n \in \mathbb{N}:f^{-1}\big(\bigcup_{\lambda=1}^{n}G_{\lambda}\big)$ is a finite subcover of $M$ and $M \subseteq f^{-1}\big(\bigcup_{\lambda=1}^{n}G_{\lambda}\big)$.
Hence, $\bigcup_{\lambda=1}^{n}G_{\lambda}$ is finite subcover of $M'$, since $f$ is bijective and has continuous inverse. 
Suppose by contradiction that $\exists f(x) \in M':f(x) \notin  \bigcup_{\lambda=1}^{n}G_{\lambda}$. Therefore, $\exists x \in M: x \notin \bigcup_{\lambda=1}^{n}G_{\lambda}$, which is a contradiction with the fact that $M$ is compact, hence, $\bigcup_{\lambda=1}^{n}G_{\lambda}$ is a finite subcover of $M'$, therefore, it is compact.
(2) Suppose $M$ is connected. Let $f:M \to M'$ be a continuous, bijective function with continuous inverse. Suppose by contradiction that $f(M) \subset M'$ is disconnected.
Therefore, exists $A \neq \emptyset$ and $B \neq \emptyset$ such as $f(M)=A\cup B$. But since $f$ is homeomorphic, $f^{-1}(A) \cup f^{-1}(B)=M$ would also be disconnected, hence, a contradiction.
I have another question: The simple fact that to be homeomorphic $f$ must be continuous wouldn't guarantee these results? 
 A: General facts: let $f: X \to Y$ be continuous.
Then $X$ compact implies that $f[X]$ is compact.
Also, $X$ connected implies that $f[X]$ is connected.
This trivially implies the homeomorphism facts as $M' = f[M]$ in your case.
A: Let me start with the final question: yes, it would be enough.
There are several errors in your proofs.
It is not true that $f^{-1}\left(\bigcup_{\lambda\in\Lambda}G_\lambda\right)$ is an open cover of $M$. It  turns out that $f^{-1}\left(\bigcup_{\lambda\in\Lambda}G_\lambda\right)=M$.
And it doesn't make sense to write about $\bigcup_{\lambda=1}^nG_\lambda$. What makes think that $1,2,\ldots,n\in\Lambda$?
Concerning connectedness, there are always sets $A$ and $B$ such that $A,B\neq\emptyset$ and that $M'=A\cup B$. That's not what connectedness means.
A: yes, all you need is that $M^{\prime}$ is a continuous image of $M$. In other words, there exists a continuous surjection $f:M \to M^{\prime}$.
for $1:$  You should consider $\bigcup f^{-1}(G_\lambda)$, a collection of open sets, rather than the preimage of the union of open sets.
you do not need the final paragraph. Just note that you found a finite subcover, and $M^{\prime} \subset f(M) \subset f(\cup f^{-1}(G_\lambda))=\cup G_{\lambda}$.
$2:$ you will need to use the fact that $A$ and $B$ are disjoint. For example, let $A,B=M^{\prime}$... then there is no contradiction to connectedness from what you've written.
