For the first part of the question:
Assume first that the norm on $U$ is given by $\lVert u \rVert_1 = \sum_{i=1}^n \lvert u_i \rvert $ where $u = \sum_{i=1}^n u_i e_i$ with respect to the basis $\{e_1,\dots,e_n\}$. With this norm we have that $\lVert u\rVert \leq 1$ implies $\lvert u_i\rvert \leq 1$. Then your estimate shows that $\lVert Mu \rVert \leq \sum \lvert u_i \rvert \lVert Me_i\rVert \leq C \lVert u\rVert_1$, where $C = \max\{\lVert Me_i\rVert \mid i=1,\dots,n\}$, so $\lVert M\rVert \leq C$.
Since $U$ is finite-dimensional, all norms on $U$ are equivalent. Therefore there is a constant $D \gt 0$ such that $\lVert u\rVert_1 \leq D \lVert u\rVert$. But then $\lVert Mu\rVert \leq C \lVert u\rVert_1 \leq CD \lVert u\rVert$, showing that $M$ is also bounded with respect to the original norm.
For the second question, use the $\varepsilon$-$\delta$-criterion for continuity. Since $M$ is continuous, it is continuous at $0 \in U$. Therefore there is $\delta \gt 0$ such that $\lVert u - 0\rVert \leq \delta$ implies $\lVert Mu \rVert = \lVert Mu - M0\rVert \leq 1$. Thus, $\lVert u\rVert \leq 1$ implies $\lVert Mu\rVert \leq 1/\delta$ and hence $\lVert M \rVert \leq 1/\delta$.
It is a general fact that the space $\mathcal{K}(X)$ of non-empty compact subsets of a separable metric space $(X,d)$ is separable with respect to the Hausdorff metric. In fact, if $D$ is a countable dense subset of $X$ then the set $\mathcal{F}(D)$ of non-empty finite subsets of $D$ is countable and dense in $\mathcal{K}(X)$.
To see this, let $K$ be any compact subset of $X$. Choose finitely many $\varepsilon$-balls with centers in $K$ whose union covers $K$ and let $x_1,\dots,x_n \in D$ be points of these balls. Then the $\varepsilon$-neighborhood of $K$ in $X$ contains $\{x_1,\dots,x_n\} \in \mathcal{F}(D)$ by construction. Conversely, the $2\varepsilon$-balls around the points $x_1,\dots,x_n$ contain $K$, so that $\delta(K,\{x_1,\dots,x_n\}) \leq 2\varepsilon$.
Now observe that sending a finite-dimensional space to its unit sphere yields an isometric embedding of the space of finite-dimensional subspaces of $X$ into $\mathcal{K}(S(X))$ where $S(X)$ is the unit sphere of the Banach space $X$. Recall that a subspace of a separable metric space is itself separable.
This approach also yields an explicit countable dense subset of the space of finite-dimensional subspaces of $X$: Choose a countable dense subset $D$ of the unit sphere of $X$ and consider the set of finite-dimensional spaces generated by finite subsets of $D$.
Best Answer
If you formulate it that generally, no, this is not true. It holds for finite dimensional vector spaces over $\mathbb{R}$ or $\mathbb{C}$. But in the field itself we already have that closed and bounded implies compact.
So if we work over the field $\mathbb{Q}$, then this is itself a one-dimensional vector space over itself, in the standard norm $|\cdot|$. And there the unit ball $[-1,1] \cap \mathbb{Q}$ is closed and bounded, but not compact: take any sequence of rationals between $[-1,1]$ that converges to an irrational number (e.g. the finite decimal partial expansions); This, or any subsequence, does not converge (as it, and any subsequence, converges in the reals to an irrational, so cannot converge to any other real, in particular no rational). So it is not sequentially compact (which is equivalent to compactness in metric spaces).
So we need the field itself to have the Heine-Borel property that closed and bounded implies compact. This does not hold for the rationals, nor (I think) for $\mathbb{Q}_p$.