Why is $(X-a)^{+}$ an increasing convex function?
Background: it is clear that for a submartingale $X:=(X_{n})_{n \in \mathbb N}$ that for any increasing convex function $\varphi$ that $(\varphi (X_{n}))_{n \in \mathbb N}$ is also a submartingale.
A textbook states as a consequence that $(X_{n}-a)^{+}$ as well as $(X_{n}\land a)$ are submartingales is which obviously implies that $(X-a)^{+}$ is an increasing convex function.
Does it necessarily follow from $X_{n} \leq E(X_{n+1}\vert \mathcal{F}_{n})$ that $X_{n} \leq X_{n+1}$ a.s.?
The claim is that $\varphi(x)=(x-a)^+ = \max\{x-a,0\}$ is an increasing convex function. This function is $$\max\{x-a,0\} = \begin{cases} x-a & x \geq a \\ 0 & x<0\end{cases}.$$ So this has nothing to do with properties of martingales, just properties of the function $\varphi:\mathbb{R}\to \mathbb{R}$.
But your second statement should say that $X \wedge a$ is a supermartingale, not a submartingale. This is because the function $\psi(x)= x \wedge a = \min\{x,a\}$ is concave.
Edit: To address your last question, no that does not follow. Consider the case where $X_1,X_2,\dots$ is simple random walk on $\mathbb{Z}$ and $\mathcal{F}_n = \sigma(X_1,\dots,X_n)$. Then $X_n \leq \mathbb{E}(X_{n+1}\mid \mathcal{F}_n)$ (in fact this $\leq$ is $=$ because SRW is a martingale). But it is not true that $X_n \leq X_{n+1}$ a.s., because it could be that $X_{n+1}=X_n -1$.