Basic facts about subsequences of a sequence $(x_n)$ $\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$
I am doing exercise 2.5.2 of Understanding Analysis by Stephen Abbott. Is my justification for the below propositions sound?

Decide whether the following propositions are true or false, providing a short justification for each conclusion.


(a) If every proper subsequence of $(x_n)$ converges, then $(x_n)$ converges as well.


(b) If $(x_n)$ contains a divergent subsequence, then $(x_n)$ diverges.


(c) If $(x_n)$ is bounded and diverges, then there exist two subsequences of $(x_n)$ that converge to different limits.


(d) If $(x_n)$ is monotone and contains a convergent subsequence, then $(x_n)$ converges.

Proof.
(a) This proposition is false. As a counterexample, consider the sequence
\begin{align*}
 \left(1,1,\frac{1}{2},1,\frac{1}{2},\frac{1}{3},1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\ldots\right)
\end{align*}
The subsequences $(1,1,1,1,\ldots)$, $\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\ldots\right)$, $\left(\frac{1}{3},\frac{1}{3},\frac{1}{3},\frac{1}{3},\ldots\right)$ are all convergent subsequences with different limits. Hence, $(x_n)$ is not a sequence.
The Monotone Convergence Theorem only guarantees: a bounded sequence of real numbers has atleast one convergent subsequence. These subsequences may or may not converge to the same limiting value.
If every proper subsequence of $(x_n)$ converges to the same limit $a$, then $(x_n)$ converges as well.
(b) This proposition is true.
Justification. If a subsequence of $(x_n)$ diverges, it implies that, given any $\epsilon >0$, there are an infinite number of real numbers in the tail that do not lie in an $\epsilon$-neighbourhood of a limit point. So, the original sequence is divergent.
(c) This proposition is true.
Justification. Assume that the sequence $(x_n)$ is bounded, but divergent. By definition, there exists large number $M$, such that $\absval{x_n} \le M$ for all $n \in \mathbf{N}$. Consider the subsequences
\begin{align*}
(x_1,x_2,x_3,x_4,\ldots) \\(x_2,x_3,x_4,\ldots) \\(x_3,x_4,\ldots) \\\vdots
\end{align*}
Each of these subsequences are bounded subsets of $\mathbf{R}$ and therefore by AoC, it is reasonable to talk about their supremum and infimum. Define the subsequences,
\begin{align*}
 u_n := \sup \{x_k:k \ge n\}\\
 d_n := \inf \{x_k:k \ge n\}
\end{align*}
$u_n$ is bounded and monotonically decreasing. $d_n$ is bounded and monotonically increasing. By the Montone Convergence theorem, $u_n$ and $d_n$ are convergent.
(d) This proposition is true.
Justification. Assume that the subsequence $(x_{n_k})$ converges to $x$. Then, given an $\epsilon > 0$, there exists $N$ such that $x_{n_k}$ lies in the interval $(x-\epsilon,x+\epsilon)$ for all $n \ge N$. Because $(x_n)$ is monotone, the values of the sequence $(x_n)$ between $x_{n_k}$ and $x_{n_{k+1}}$ also lie in $(x-\epsilon,x+\epsilon)$. Consequently, the same $N$ can be used for the original sequence.
 A: For a) I think it's true.
First you counter-example does not work : it has many non-convergent subsequences, for example $(1, 1/2, 1, 1/2, ...)$.
Then it's actually true even if you only suppose that $(x_{2n})$, $(x_{2n+1})$ and $(x_{3n})$ converge. In fact, then if we note $a, b, c$ the limit of each, we have $(x_{6n})$ that can be extracted from both $(x_{3n})$ and $(x_{2n})$, so $a=c$, and considering $(x_{6n+3})$, we also have $b=c$. So $a=b$.
Thus $(x_n)$ converges.
For b) you can also proceed by contraposition : if $(x_n)$ converges, then all subsequences converge.
For c), as $(x_n)$ is bounded, we can extract $(x_{\varphi(n)})$ that converges. Let's call $\ell$ its limit. As $(x_n)$ does not converge to $\ell$, there exists an $\varepsilon$ such that for all $N\in \mathbb{N}$, there exists a $n \geqslant N$ such that $|x_n - \ell| > \varepsilon$. So we can extract another subsequence that does not converge to $\ell$. This subsequence being bounded too, we can extract from it a another subsequence that converges to $\ell ' \neq \ell$. So we found two subsequences that converge to different limits.
For d), you proof is fine.
