Let $W=\lbrace z\in \mathbb{C}\mid 2 \leq |z| \leq 3 \rbrace$. Prove that $W$ is connected and compact.
Let $z=x+iy,\: x,y\in\mathbb{R}$. Notice that $$W=\lbrace z\in \mathbb{C} \mid 4 \leq x^2+y^2 \leq 9\rbrace.$$ Now consider $$U=\lbrace z\in \mathbb{C}\mid x^2+y^2 \leq 9\rbrace ,\, \, V=\lbrace z\in \mathbb{C}\mid x^2+y^2 \geq 4 \rbrace. $$
Note that $U$ is closed and $\mathbb{R}^2\setminus V$ is open, hence $V$ too closed and $W=U \cap V$ hence $W$ is closed since is finite intersection of closed sets.
Now is pretty easy see that for any $z\in W$ we have the following inequality holds $|z| \leq 9$, it show that $W$ is bonded and hence $W$ is compact.
Now for see that the set is path connected I think that I should prove that $W$ is path connected. I dont´t have a formal proof the unique idea that comes to my mind is:
Let $w_1,w_2\in W$ there are two cases.
Case I
If $|w_1|=|w_2|$ then we only need consider the circunference with center in $(0,0)$ and radious $|w_1|$ let us denote by $C(0,|w_1|)$ is easy see that $C(0,|w_1|)\subset W$ hence the arc of circunference that connect $w_{1}$ with $w_2$ is our path.
Case II
If $|w_1|\neq |w_2|$.Then consider the straight line that join $w_1$ with $w_2$ let us denote by $\ell$ Then consider $m=Min\lbrace|w_1|,|w_2| \rbrace $. WLOG assume that $m=|z_1|$ and consider $C(0,m)$ notice that $\ell \cap C(0,m)\neq \emptyset $ since it contains at most the complex number with norm $m$.Frome here since $\ell$ is a straight line must intersect $C(0,m)$ in a point $P\neq w_1$.
WLOG let us take the arc of circunference from $w_1$ to $P$ and the path from $P$ to $w_2$ which is contained in $W$(dangerous argue),hence the union of both path is the path requiered.
Is my answer right? Is possible find a more formal proof of the set $W$ is connected? In my proof I have a due in a argue: Is always true that the path from $w_1$ to $P$ and the path from $P$ to $w_2$ is always contained in $W$?
Your argument is correct for Case I but not for Case II. Your point $P \neq w_1$ many not exist.
Consider $C(0,|w_1|)$ and show that there is a point $P$ on this which is a positive multiple of $w_2$. To be precise, the point is $w_1e^{it}$ where $t$ is the argument of $\frac {w_2} {w_1}$. Now go along the circle $C(0,|w_1|)$ from $w_1$ to $P$ and then go to $w_2$ using the line segment from $P$ to $w_2$.