Nice problem

If $f$ is continuous, consider $f$ as a map from $[0,1]$ to $\mathbb{R}$. The domain is compact, so its image is compact, and therefore closed in $\mathbb{R}$, and this cannot be $(0,1)$, since it is not closed, therefore any continuous map from $[0,1]$ to $(0,1)$ cannot be surjective.

b. Let $(y_n)$ be a sequence in $(0,1)$ converging to $1$. By surjectivity, we can find a sequence $(x_n)$ in $[0,1]$ such that $f(x_n) = y_n$. $(x_n)$ admits a convergent subsequence $(x_{n_m})$ converging to some $x \in [0,1]$. Then we have $f(x_{n_m}) \to 1$, but $f(x)$ cannot be $1$, as $1$ is not in the range. Therefore $f$ is discontinuous as $x$. The same argument can be applied to a sequence $(y'_n) \subseteq (0,1)$ converging to $0$ to produce a convergent (sub)sequence $(z_{n_m}) \subseteq [0,1]$, converging to a point $z \in [0,1]$ where $f$ is also discontinous. To show $x$ and $z$ are different, and therefore that there are two separate discontinuities, suppose that they were the same. Then $ x=z$ are limit points of subsequences $(x_{n_m})$ and $(z_{n_m})$ where $\lim_{m \to \infty} f(x_{n_m}) = 1$ and $\lim_{m \to \infty} f(z_{n_m}) = 0$, which would mean that $f$ does not admit a limit at the point $x=z$, since the limit $\lim_{w \to x} f(w)$ does not exist. But this is against hypothesis.

Is there any easy proof for (a) without using compactness?