Let $(X,\| \cdot \|_X)$ be a $\mathbb{R}$-vector space with bidual space $X^{**}$. We defined $X$ to be reflexive, if the canonical embedding $\mathcal I: X \to X^{**}$ with $$\mathcal I x(l) := l(x)$$ for all $l \in X^*, x \in X$ is surjective. I know for example Hilbert spaces are reflexive, $L^p$ spaces are reflexive for $1 < p < \infty$ and $L^1$ is not reflexive. However, I have no idea how I should think about reflexive spaces. I have seen theorems requiring reflexivity in the proof, however I don't understand why there is no proof if the canonical embedding is not surjective (of course I can't find any such proof either).
For example, we have seen the following "approximation theorem": Let $X$ be reflexive, $M \subset X$ non-empty, convex and closed, $x_0 \in X \setminus M$. Then, there exists $m_0 \in M$ such that $$\| x_0 - m_0 \|_X = \operatorname{dist} (x_0, M) = \operatorname{inf}_{m \in M} \| x_0 - m \|_X.$$ It feels to me as if the fact that $M$ is closed should suffice for this theorem, however of course I am imagining the space as $\mathbb{R}^n$ which I guess I shouldn't.
Is there some "property" every reflexive space automatically gains? I only see that we can use surjectivity very efficiently to prove those theorems, however I have no clue why there can't be any other proof without surjectivity or without using the canonical embedding at all. Usually, for other requirements of a theorem, it is easy to understand why certain property is crucial but with this property I am struggling.