Using the definition that a function $f : A \to \mathbb{R}\cup\left\{ -\infty, \infty \right\}$ is l.s.c at a point $\bar{x} \in A$ if $\forall \alpha < f( \bar{x} ), \exists \delta > 0$ s.t. $\forall x \in ( \bar{x}-\delta , \bar{x} + \delta ), \alpha < f( x )$, show that $ f(x) = \begin{cases} x & \text{if } x \leq 4 \\ x + 1 & \text{if } x > 4 \end{cases}$ is lsc at $\bar{x} = 4$.
Following the definition, for any $\alpha < f(\bar{x}) \to \alpha < 4$ I need to find some $\delta$-neighborhood of $\bar{x}$ (here, open interval) contaning points $x$ that verify $\alpha < f( x )$.
I reckon it's possible, but I only see how it fails: Any neighborhood here has a point $x_{0}<\bar{x}$ as $\delta >0$, so $f( x_0 ) < f( \bar{x} )$ and $\alpha < f( x )$ cannot be verified for all $\alpha < f( \bar{x})$. For example $\alpha = 3 > f( 2.5 )=2.5$ assuming $\delta = 2$. There might be equality that $\alpha = f( x )$ when interval is as narrow as possible, but I don't see how $\alpha < f( x )$ can hold.
That's why it's just "there exists $\delta"$ which will depend on $\alpha$. For example, take $\alpha=3$, then I can take $\delta=0.5$. The order in which you pick the variables is key: I fix $\alpha$ first then find a $\delta$ that works based on that $\alpha$.
More generally, fix $\alpha<f(4)=4$. Then choose any $\delta$ such that $0<\delta<4-\alpha$ (e.g. $\delta:=\frac{4-\alpha}{2}$). Then $4-\delta>\alpha$, so for $x\in (4-\delta,4+\delta)\subseteq (\alpha,\infty)$ we have $$f(x)=\begin{cases}x>\alpha \qquad\qquad\qquad~~~~~ x\leq 4 \\ x+1>\alpha+1>\alpha \qquad x>4\end{cases}$$ so $f(x)>\alpha$.