How can I show that this shape is connected Is the following shape connected? My intuition says yes...but I cannot prove  it...any ideas please? $$\left\lbrace(x,y)\in\mathbb{R^2}:{x^2+y^2=1,0<x\le1}\right\rbrace\cup\left\lbrace{(x,y)\in\mathbb{R^2}: 0<x\le1,y=-1}\right\rbrace$$
 A: Tsemo appears to have answered the question in the comments beneath his post, so I'll just slightly generalize and summarize here to record a complete answer.
Suppose $A$ and $B$ are nonempty, connected subsets of a topological space $X$. Then $A\cup B$ is disconnected if and only if $\bar{B}\cap A = \bar{A}\cap B = \newcommand{\nullset}{\varnothing}\nullset$.
Proof: First suppose the latter condition holds. We'll prove this implies $A\cup B$ is disconnected. $X\setminus \overline{B}$ is open in $X$, so $(X\setminus\overline{B})\cap (A\cup B)$ is relatively open in $A\cup B$.
But 
$$(X\setminus\overline{B})\cap (A\cup B)
=(A\setminus\overline{B})\cup (B\setminus\overline{B})
=A\cup\nullset
= A.
$$
Hence $A$ is relatively open in $A\cup B$. Symmetrically $B$ is relatively open in $A\cup B$. Hence $A$ and $B$ disconnect $A\cup B$. (This is essentially the content of Tsemo's comment on his answer.)
Now suppose $A\cup B$ is disconnected, say by open sets $U$, $V$ of $X$ with $U\cap V \cap (A\cup B)=\nullset$, and $(U\cup V)\cap (A\cup B)=A\cup B$. Then $U\cap A$ and $V\cap A$ would disconnect $A$ if both were nonempty, but we assumed $A$ is connected, hence one of these is empty, and the other contains $A$. Suppose WLOG that $A\subset U$ and $A\cap V=\nullset$. Then similarly, since $B$ is connected, and we can't have $V\cap (A\cup B)=\nullset$, we must have $B\subset V$ and $B\cap U=\nullset$. Hence $B\subseteq U^C$, and $U^C$ is closed and disjoint from $A$. Thus $\overline{B}\cap A =\nullset$. Symmetrically, $\overline{A}\cap B = \nullset$.
The application to your question:

As for your case, the only point of intersection between the closures of the semicircle and the line segment is $(0,-1)$. This is not contained in either the semicircle or the line segment, hence the closure of the semicircle is disjoint from the line segment and vice versa. Thus the union is disconnected.

A: draw a picture, it is not connected since it is the union of the half of a circle and a segment and they are disjoint.
