Let $\overline{\mathbb{R}}=\mathbb{R}\cup\{+\infty,-\infty\}$. For a secuence $\{x_n\}_{n\in\mathbb{N}}\subseteq \overline{\mathbb{R}}$, the inferior limit of it is defined as
$$\operatorname{lim inf}_{n\to\infty}(x_n):=\operatorname{sup}\{\operatorname{inf}(\{x_n\}_{n\geq m})|m\in\mathbb{N}\}.$$
The proposition is the following:
$$\operatorname{lim inf}_{n\to\infty}(x_n)=\operatorname{min}\{z\in \overline{\mathbb{R}}|z \text{ is a subsequential limit of } (x_n)_{n\in\mathbb{N}}\}.$$
The reason I'm asking this is that according to other sources, the actual result is $$\operatorname{lim inf}_{n\to\infty}(x_n)=\operatorname{inf}\{z\in \overline{\mathbb{R}}|z \text{ is a subsequential limit of } (x_n)_{n\in\mathbb{N}}\}$$ but I cannot seem to find any mistake in the proof given by my professor, whose sketch I leave in the following lines.
Sketch of proof:
For each $m\in\mathbb{N}$, let $y_m=\operatorname{inf}\{x_n\}_{n\geq m}$. It follows imnmediately that $y_m\leq y_{m+1}$ for each $m$.
We divide the proof in three cases.
Case 1. $\operatorname{lim inf}_{n\to\infty}(x_n)=+\infty$.
Let $M\in \mathbb{R}$ be an arbitrary number. Since $M<+\infty=\operatorname{lim inf}_{n\to\infty}(x_n)$, we have the existence of an $m\in \mathbb{N}$ such that $M<y_m$. Then, given that $y_m=\operatorname{inf}\{x_n\}_{n\geq m}$, we have that for each $n\geq m$, $M<y_m\leq x_n$ so $M<x_n$. Since $M$ was arbitrary, the existence of such $m$ guarantees that $\lim_{n\to\infty}(x_n)=+\infty$ and therefore any subsequence of $(x_n)_{n\in\mathbb{N}}$ converges to $+\infty$.
It is then evident that $\operatorname{lim inf}_{n\to\infty}(x_n)=\operatorname{min}\{+\infty\}=\operatorname{min}\{z\in \overline{\mathbb{R}}|z \text{ is a subsequential limit of } (x_n)_{n\in\mathbb{N}}\}.$
Case 2. $\operatorname{lim inf}_{n\to\infty}(x_n)=-\infty$.
We have that for each $m$, $y_m\leq \operatorname{lim inf}_{n\to\infty}(x_n)=-\infty\leq y_m$, so in fact, $y_m=-\infty$. Then, given any $m$ we have $-\infty=y_m=\inf\{x_n|n\geq m\}$. Therefore, for any $M\in \mathbb{R}$, it is true that $-\infty <M$, which indicates the existence of an index $n \geq m+1>m$ such that $x_n<M$. Let $k^{(M,m)}=\min\{n\in\mathbb{N}|m<n\land x_n<M\}$ and define recursively $n_0:=k^{(0,0)}$ and having defined $n_m$, define $n_{m+1}:=k^{(x_{n_m},n_m)}$ in case $-\infty<x_{n_m}$ and if $x_{n_m}=-\infty$ simply define $n_{m+1}:=n_m+1$. It is not hard to see that in fact $(x_{n_m})_{m\in\mathbb{N}}$ is a subsequence of $(x_n)_{n\in\mathbb{N}}$ that converges to $-\infty$ by construction. It follows that any subsequential limit of the sequence must be greater or equal than $-\infty$, which is in itself a subsequential limit and it is possible now to conclude that $\operatorname{lim inf}_{n\to\infty}(x_n)=\operatorname{min}\{z\in \overline{\mathbb{R}}|z \text{ is a subsequential limit of } (x_n)_{n\in\mathbb{N}}\}.$
Case 3. $\operatorname{lim inf}_{n\to\infty}(x_n)\in\mathbb{R}$. For convenience, let us write $x^*=\operatorname{lim inf}_{n\to\infty}(x_n)$. The first part of the argument is proving that for any $z<x^*$, $z$ cannot be a subsequential limit of $(x_n)$ and therefore $x^*$ is a lower bound of the set of subsequential limits. Then, one notes that for any $\varepsilon>0$ and any $N\in\mathbb{N}$, one can find $n>N$ such that $x_n\in(x^*-\varepsilon,x^*+\varepsilon)$:
Take $z<x^*=\sup\{y_m|m\in\mathbb{N}\}$. There must exist an index $m\in\mathbb{N}$ such that $z<y_m$, but $y_m=\inf\{x_n|n\geq m\}$ so we then have $z<y_m\leq x_n$ for any $n\geq m$. If $z=-\infty$ then $[z,y_m)$ is an open neighbourhood of $z$ that contains at most a finite amount of terms of the sequence (meaning $\{k\in\mathbb{N}|x_k\in [z,y_m)\}$ is a finite set) and therefore $z$ cannot be a subsequential limit. If $z\in \mathbb{R}$, then $z<y_m\leq x^*\in\mathbb{R}$ implies $y_m\in \mathbb{R}$ and so if we pick $\varepsilon>0$ such that $z+\varepsilon<y_m$ then $(z-\varepsilon,z+\varepsilon)$ is an open neighbourhood of $z$ that contains at most a finite amount of terms of the sequence and hence $z$ itself is not a subsequential limit. No matter the case, $z$ is not a subsequanetial limit of $(x_n)$ and we can conclude that $x^*$ is a lower bound of the set of subsequential limits of $(x_n)$.
Given $\varepsilon>0$ and $N\in\mathbb{N}$, we shall give an $n>N$ such that $x_n\in (x^*-\varepsilon,x^*+\varepsilon)$. Since $x^*-\varepsilon<x^*=\sup\{y_m|m\in\mathbb{N}\}$, we can guarantee the existence of an index $M_1\in\mathbb{N}$ such that $x^*-\varepsilon<y_{M_1}\leq x^*$. Now, pick $N^\#>\max\{N,M_1\}$. Since $(y_m)$ is a non-decreasing sequence, we then have $x^*-\varepsilon<y_{M_1}\leq y_{N^\#}\leq x^*<x^*+\varepsilon$, but then $\inf\{x_k|k\geq N^\#\}=y_{N^\#}<x^*+\varepsilon$ so there must be an index $n\geq N^\#>N$ such that $x^*-\varepsilon<y_{N^\#}\leq x_n<x^*+\varepsilon$ which in turn implies $x_n\in (x^*-\varepsilon,x^*+\varepsilon)$ and $n>N$.
With the past points proved, it is now true that $x^*$ is a lower bound of the set of subsequential limits of $(x_n)$ and that one can construct, in a similar manner to that of case 2 a subsequence of $(x_n)$ that converges to $x^*$, so $x^*$ is itself a subsequenial limit and hence, $\operatorname{lim inf}_{n\to\infty}(x_n)=\operatorname{min}\{z\in \overline{\mathbb{R}}|z \text{ is a subsequential limit of } (x_n)_{n\in\mathbb{N}}\}.$
The proposition is now proved for all possible cases.