Solution:
(1)
Let $[f(c_i)]$ be a sequence in $f([a,b])$, and suppose it tends to $y \in \mathbb{R}$. Then the $c_i$ are elements of the closed bounded interval $[a,b]$, so they have a convergent subsequence $c_{n_i}$, say; by closure of $[a,b]$, the limit $x$ of the $c_{n_i}$ lies in $[a,b]$.
Now since $c_{n_i} \to x$, have $f(c_{n_i}) \to f(x)$ by continuity of $f$, so the limit of $[f(c_{n_i})]$ lies in $f([a,b])$.
Finally, since $[f(c_{n_i})]$ is a subsequence of the convergent-by-assumption $[f(c_i)]$, it converges to the same limit; so $f(x) = y$.
This proves that $f([a,b])$ contains all its limit points.
Can somebody help me with (2)
Best Answer
For 2): $y_n \to y$. If $(a_n)$ is a subsequence of $(x_n)$ which converges to some point $t$ then $f(a_n)$ is a subsequence of $f(x_n))=(y_n)$ so it converges to $y$. But $f(x_n) \to f(t)$, so $f(t)=y=f(x)$. Since $f$ is injective we get $t=x$, so $x_n \to x$. Thus any subsequence of $(x_n)$ which is convergent has limit $x \,\,$ (1). Now suppose $(x_n)$ does not converge to $x$. Then there exists $\epsilon >0$ and $n_1<n_2<\cdots$ such that $|x_{{n_k}}-x| >\epsilon $ for all $k$. But $(x_{{n_k}})$ is a bounded sequence of real numbers sio it has a convergent subseqeunce $(a_n)$. The inequality $|x_{{n_k}}-x| >\epsilon $ gives $|a_n-x| >\epsilon $ for all $n$ which contradicts $\,\,$ (1).