Which of the following statements are true? (NBHM 2017)
$a.$ Let $X$ be a set equipped with two topologies $\mathcal{T}_1$ and $\mathcal{T}_2$. Assume that any given sequence in $X$ converges with respect to the topology $\mathcal{T}_1$ if, and only if, it also converges with respect to the topology $\mathcal{T}_2$. Then $\mathcal{T}_1$ = $\mathcal{T}_2$.
$b.$ Let ($X$; $\mathcal{T}_1$) and ($Y$; $\mathcal{T}_2$) be two topological spaces and let $f$ : $X \to Y$ be a given map. Then $f$ is continuous if, and only if, given any sequence {$x_n$}n$\epsilon_\Bbb{N}$ such that $x_n \to x$ in X, we have $f(x_n) \to f(x)$ in $Y$ .
$c.$ Let ($X$; $\mathcal{T}$) be a compact topological space and let {$x_n$}n$\epsilon_\Bbb{N}$ be a sequence in $X$. Then, it has a convergent subsequence.
Answer: all of them are false.
I came up with counterexamples for ($b$) and ($c$), but I'm unable to find one for ($a$).
For ($b$) I assumed, $X=Y$ and $\mathcal{T}_1$ as discrete topology and $\mathcal{T}_2$ as indiscrete topology, and $f$ :$X\to$$Y$ as identity map.
Now because of topological properties of discrete and indiscrete topologies, $f$ is continuous. If we assume constant sequence in $X$ ( which converges). then its image under $f$ in $Y$ will be a constant sequence, but the constant sequence does not converge in indiscrete topology. Hence $(b)$ is false.
For $(c)$ assuming ($X$; $\mathcal{T}$) as indiscrete topology (Indiscrete topology is compact for any space $X$) and {$x_n$}n$\epsilon_\Bbb{N}$ be constant sequence, whose any subsequence will be a contant sequence, and again we know that constant sequence does not converge in indiscrete topology. Hence $(c)$ is also false.
But I'm unable to find counterexample for $(a)$
Please help. Thanks in advance.
EDIT: As suggested by David Hartley, in indiscrete topology every sequence converges to each point of the space, my arguments for $(b)$ and $(c)$ will no longer hold true, so I'm also requesting for the counterexamples for ($b$) and ($c$). ( I'm also trying to find out)
