If you want to prove an inequality in real analysis there are two main ways of doing that.
First, we could show that $a \le b + \varepsilon$ for every $\varepsilon > 0$.
The quantity $b + \varepsilon$ should remind you of the interval $(b - \varepsilon, b + \varepsilon)$. We know that $b_n \to b$ means that for every $n \ge $ some $N$, we have $b_n \in (b - \varepsilon, b + \varepsilon)$. So in particular, $b_n \le b + \varepsilon$ and therefore $a_n \le b_n \le b + \varepsilon$ (this is for $n \ge N$).
This is a step in the right direction because we have replaced $a_n \le b_n$ with $a_n \le c$ where $c = b + \varepsilon$ is constant which, at least in principle, is simpler.
Second, we could argue by contradiction, as you have done.
Suppose that $a > b$, which one should think of as $a = b + \varepsilon$ where $\varepsilon = a - b > 0$ and proceed as before. We know that for $n \ge N$ we have $b_n \in (b - \varepsilon, b + \varepsilon)$ so $b_n \le b + \varepsilon = a$ and hence $a_n \le b_n \le a$.
You will notice that this isn't particularly helpful. So we try again. What happens if we make $\varepsilon$ smaller, say to $\varepsilon = \frac{a - b}{2}$? Well, then we have
$$ a_n \le b_n \le b + \varepsilon = b + \frac{a - b}{2} \tag{$*$} $$
Again we have something of the form $a_n \le$ some constant. So if we appeal to the theorem that says that if $a_n \le c$ (a constant) and $a_n \to a$ then $a \le c$ we have from $(*)$:
$$ a \le b + \frac{a - b}{2} < b + (a - b) = a, $$
and this is a contradiction.
As an exercise, you might like to finish the first method as well.
Let $$c=\liminf (a_n+b_n),\quad a=\liminf a_n,\quad b=\liminf b_n$$ There exists a subsequence $a_{n_k}+b_{n_k}$ convergent to $c.$ Since $a_{n_k}$ is bounded, we can choose a subsequence $a_{n_{k_l}},$ which is convergent, say to $\tilde{a}.$ Hence $\tilde{a}\ge a.$ Next as $b_{n_{k_l}}$ is bounded, we can choose a subsequence $b_{n_{k_{l_m}}},$ which is convergent, say to $\tilde{b}.$ Then $\tilde{b}\ge b.$ Moreover $a_{n_{k_{l_m}}}\to \tilde{a}. $ Therefore
$$c=\lim_{m\to\infty}\left [a_{n_{k_{l_m}}}+b_{n_{k_{l_m}}}\right ]=\tilde{a}+\tilde{b}\ge a+b$$
Best Answer
Part (i) looks good. For part (ii), your general idea is right, but you have a small error in your definition of what you have to show:
Actually you should show that if $b \le a_n$ for all $a_n \in \mathbb{R}$, then $b \le 0$. (So $b$ could be negative). Remember that infimum = "greatest lower bound". So you want to show that for any lower bound $b$ of $a_n$, that lower bound is less than or equal to $0$.
Then in your reasoning, you can't assume $0 < b$. It looks like you have currently assumed that here:
But if you replace this with $b \le a_n < b_n$ (just remove "$0 < $"), and apply the same idea, you should get the conclusion you want.