A function is lower semicontinuous at $x \in\mathbb{R}$ if given any sequence $x_n$ converging to $x$, $$f(x)\leq\lim_{n \to \infty} f(x_n)$$
Provided that the limit on the right exists.
Let $$I_{u}(x) = \begin{cases} 1, & \quad x > 0 \\ 0, & \quad x \leq 0 \end{cases}$$
Show $I_u(x)$ is lower semicontinuous.
Attempt:Let $x_n$ be any sequence that converges to $0$.Then $\lim_{n \to \infty} I_u(x_n)=0$ or $\lim_{n \to \infty} I_u(x_n)=1$.
In either case $I_u(0)=0\leq \lim_{n \to \infty} I_u(x_n)$.
Is this argument correct or do I have to show why $\lim_{n \to \infty} I_u(x_n)=0$ or $\lim_{n \to \infty} I_u(x_n)=1$?
According to the definition, your argument is correct. It would make the argument more complete if you, as mentioned, added why only $0$ and $1$ are the limits if they exist (although obvious).
As for the definition, you will have to be careful in applying it further. Continuity at $x$ for $f$ in terms of sequences is that for any sequence $(x_{n})$ with $x_{n}\to x$ we have $f(x_{n})\to f(x)$, but, if you define upper semi-continuity in a similar way then for a upper and lower semi-continuous function $g$ there could exist a sequence $(y_{n})$ with $y_{n} \to y$ but $g(y_{n})$ does not converge to $g(y)$. That is, $g$ would not be continuous.