A subspace of a complete metric space is compact iff it is closed and totally bounded.
Take a compact space $X$, then its space of continuous functions $C(X)$ is complete. Subspaces of $C(X)$ are compact iff they are closed, bounded and equicontinuous.
For such subspaces, the above two theorems show that closed, bounded and equicontinuous is equivalent to closed and totally bounded.
Can we "cancel" closedness in the above equivalence so we have bounded and equicontinuous iff totally bounded?
Also, we know that totally bounded implies bounded. Does that mean that totally bounded implies equicontinuous in the case of such spaces?
Yes, you can "cancel" closedness. In fact that's often exactly how the Arzela-Ascoli theorem is stated. The statement of Arzela-Ascoli that is taught at my course is as follows:
Let $K$ be a compact space and $C(K)$ be the space of continuous real valued functions on $K$. Then $S \subset C(K)$ is totally bounded if and only if it is bounded and equicontinuous.
In a similar vein rather than the result in your first sentence a slightly different result that talks about non-closed $S$ is: a subset $S$ of a complete metric space is pre-compact (meaning $\bar{S}$ is compact) if and only if $S$ is totally bounded. It's equivalent to the result you've stated since $S$ is totally bounded if and only if $\bar{S}$ is totally bounded.
Finally your second bullet point is correct, a totally bounded subset of $C(K)$ is equicontinuous since it's an if and only if statement.
Proof of Arzela-Ascoli:
Firstly let $S$ be totally bounded. Then certainly $S$ is bounded. To show equicontinuity, fix a point $x \in K$ and $\epsilon > 0$. Then there exists an $\epsilon$-net $\{f_1, \ldots, f_n\} \subset S$. Then since each $f_i$ is continuous there exist an open neighbourhood of $x$, $U_i$, such that for all $y \in U_i$ we have $\lvert f_i(y) - f_i(x) \rvert < \epsilon$.
Now take $U = \cap_{i=1}^n U_i$ which is still an open nbhd of $x$. Then for any $y \in U$ and $f \in S$, there exists $f_i$ such that $\lVert f - f_i \rVert < \epsilon$. Then $$ \lvert f(y) - f(x) \rvert \leq \lvert f(y) - f_i(y) \rvert + \lvert f_i(y) - f_i(x) \rvert + \lvert f_i(x) - f(x) \rvert < 3 \epsilon $$ So $S$ is equicontinuous.
Conversely let $S$ be bounded and equicontinous. Then by equicontinuity for all $x \in K$ ther exists an open nbhd $U_x$ of $x$ such that for all $y \in U_X$, $f \in S$ we have $\lvert f(x) - f(y) \rvert < \epsilon$. Then $K = \cup_{x \in K} U_x$ so by compactness $K = \cup_{i = 1}^n U_{x_i}$ for some $x_1, \ldots, x_n$. Now consider the map $$ \varphi: C(K) \to \mathbb R^n, \quad \varphi(f) = (f(x_1), \ldots, f(x_n))$$ It is easy to check that $\varphi$ preserves boundedness of $S$, thus $\varphi(S)$ is a bounded subset of $\mathbb R^k$. Then by Heine-Borel $\overline{\varphi(S)}$ is a compact, thus $\overline{\varphi(S)}$ and hence $\varphi(S)$ is totally bounded in $\mathbb R^k$. Now fix $\epsilon > 0$ and pick an $\epsilon$-net $\{\varphi(f_1), \ldots, \varphi(f_n)\}$ of $\varphi(S)$ in $\mathbb R^k$, we show $\{f_1, \ldots, f_m\}$ is an $3\epsilon$-net of $S$ in $C(K)$.
Fix any $y \in K$ and $f \in C(K)$. Then there exists $f_j$ such that $\lvert f(x_i) - f_j(x_i) \rvert < \epsilon$ for all $i = 1, \ldots, n$. Then $K = \cup_{i=1}^n U_{x_i}$ thus $y \in U(x_i)$ for some $i$. Then $$ \lvert f(y) - f_j(y) \rvert \leq \lvert f(y) - f(x_i) \rvert + \lvert f(x_i) - f_j(x_i) \rvert + \lvert f_j(x_i) - f(y) \rvert < 3\epsilon$$ Thus $\lVert f - f_j \rVert < 3\epsilon$ so as required $\{f_1, \ldots, f_m\}$ is a $3\epsilon$-net of $S$. So $S$ is totally bounded.