let's say we have a partially defiend function
$$ f(x)=\left\{ \begin{align} x+1 & \quad\mbox{if}\quad x<5\\ x-5 & \quad\mbox{if}\quad x\ge5 \end{align} \right. $$
The graph of that function would look like this: the graph of function from geogebra In this specific case,would x = 5 be local minima?
As always, if we have a question like this we want to go back to the Definition. I'll rewrite it just a little bit, to make it more applicable:
A real number $x_0$ is a local minimum of a real function $f\colon\mathbb{R}\to\mathbb{R}$ if there exists $\epsilon>0$ such that for all real $x$, $|x-x_0|<\epsilon\implies f(x_0)\leq f(x)$. In other words, $x_0$ is a local minimum if $f(x_0)$ is the smallest of all $f(x)$ in some neighborhood around $x_0$.
In this case $5$ is indeed a local minimum, because $f(5+\epsilon)>f(5)$ for all positive real $\epsilon$, and $f(5-\epsilon)>f(5)$ for all positive real $\epsilon<5$.
However, it's important to note that even though local extrema do not require differentiability or even continuity, being a minimum must be "two-sided" in some sense; i.e. the function
$$\begin{cases}4-x,&x<5\\x-5,&x\geq5\end{cases}$$
does not have a local minimum at $5$, because $f(5)$ is greater than $f(4.9),\ f(4.99),\ f(4.999),\ $etc.