Question about limit inferior for infinite cases

Question

Let $(X,d)$ a metric space, $f$ is a function from $X$ to $\mathbb{R}$, and $(x_n)$ is a sequence which converges to $a$, then $$\liminf_{x \to a} f(x)\leq \liminf_{n\to\infty} f(x_n)$$ I have done for the case that the values are real here. Now, I realize that the value of limit inferior can be $-\infty$ or $\infty$. So, I think I need to prove for additional cases, i.e:

1. If $\liminf_{x\to a} f(x)=\infty$,
2. If $\liminf_{x\to a} f(x)=-\infty$,
3. If $\liminf_{n\to \infty} f(x_n)=\infty$,
4. If $\liminf_{n\to \infty} f(x_n)=-\infty$,

I try to prove for the cases above, here is my attempt:

1. If $\liminf_{x \to a} f(x)=\infty$, then $\infty \leq \liminf_{n\to\infty} f(x_n).$ Because there is no real number which is greater than $\infty$, so we must have $\liminf_{n\to\infty} f(x_n)=\infty$. Therefore, $\liminf_{x \to a} f(x)\leq \liminf_{n\to \infty} f(x_n)$.
2. If $\liminf_{x \to a} f(x)=-\infty$, diperoleh $-\infty\leq \liminf_{n\to \infty} f(x_n).$ Suppose that $\liminf_{n\to \infty} f(x_n)=-\infty$, then $\liminf_{x \to c}f(x)=\liminf_{n\to \infty}f(x_n)$ and if $\liminf_{n\to \infty} f(x_n)$ is real or $\infty$, then $-\infty<\liminf_{n\to \infty} f(x_n)$. Therefore, the inequality is true.
3. Similar to case (2).
4. Similar to case (1).

Is my attempt correct?
Thanks for any advice.