Let's sketch the graph of this function first over $[-1,1]$.
It looks messy:
$f(x)= \cases{\color{darkgreen}{0}, & $\color{darkgreen}{x=-1}$ \cr\color{maroon}{ -x }& $\color{maroon}{-1\lt x\lt 0} $\cr \color{darkblue}{x}, & $\color{darkblue}{ {0\le x\lt 1} }$ \cr}$
$f$ s a piecewise defined function. To sketch the graph of $f$, look at each "piece" separately:
When $\color{darkgreen}{x=-1}$, the function has the value $0$; so the point $(-1,0)$ is on the graph. We plot this point in the graph below.
For the range of $x$-values in $\color{maroon}{-1
For the range of $x$-values in $\color{darkblue}{0\le x\le1}$, the rule for $f$ is
$\color{darkblue}{f(x)= x}$. Over the interval $(0,1)$ sketch the graph of $\color{darkblue}{y=-x}$. Note that the value of $f$ at $ 1$ is 1.

So, that's the graph of $f$ over $[-1,1]$.
What about the rest of the graph? Well, that's obtained from the rule
$$
f(x+2)=f(x)+1.
$$
So, for example: $f(3)=f(1+2)=f(2)+1$ and $f(3.5)=f(1.5+2)=f(1.5)+1$.
To evaluate $f$ at $f(x+2)$, you just take the value of $f$ at $x$, and add 1.
So, to obtain the graph of $f$ over $[1,3]$, you just draw the same shape as the graph over $[1-2, 3-2]=[-1,1]$ but raise it vertically by 1 unit.
To obtain the graph of $f$ over $[3,5]$, you just draw the same shape as the graph over $[3-2, 5-2]=[1,3]$ but raise it by 1 unit. etc...
Below, is the graph of $f$ over three of its "periods":
Now, looking at the graph, you can see that $f$ discontinuous at every odd integer $x\ge -1$ and that $f$ is continuous for any $x>-1$ that is not an odd integer.