Suppose $\mathcal{L}$ is a finite-dimensional Lie algebra, and $\mathcal{G} = e^{\mathcal{L}}$ is it's compact, connected Lie group.
Given a closed sub-algebra $\mathcal{L}' \subset \mathcal{L}$, it naturally follows that $\mathcal{G}' = e^{\mathcal{L}'}$ is a closed subgroup of $\mathcal{G}$.
Does the converse also hold?
That is, for two compact, connected Lie groups $\mathcal{G} = e^{\mathcal{L}}, \mathcal{G}' = e^{\mathcal{L}'}$ where $\mathcal{G}' \subset \mathcal{G}$, does it follow that $\mathcal{L}' \subset \mathcal{L}$?
I suspect that the answer is 'no', but I don't have the foggiest how to prove it.
Any help would be appreciated.
It seems as though you have a few misconceptions.
First, given a finite dimensinoal Lie algebra $\mathfrak{g}$, there may exist a compact, connected Lie group $G$ with Lie algebra $\mathfrak{g}$. If there does exist such a Lie group, there is no reason it must be unique. If one adds the hypothesis "simply connected" or "centerless", then existence implies uniqueness.
For example, the Lie algebra $\mathfrak{sl}(2,\mathbb{R})$ is not the Lie algebra of any compact Lie group. The Lie algebra $\mathfrak{su}(2)$ is the Lie algebra of the nonisomorphic compact, connected Lie groups $SU(2)$ and $SO(3)$. ($SU(2)$ is simply connected and $SO(3)$ is centerless)
Second, given a subalgebra $\mathfrak{h}\subseteq \mathfrak{g}$, it is always closed (if $\mathfrak{g}$ is finite dimensional.) If $G$ has Lie algebra $\mathfrak{g}$, then there is always a unique connected subgroup $H\subseteq G$ with Lie algebra $\mathfrak{h}$, but often, $H$ is not a closed subgroup. For example, let $\mathfrak{g} = \mathbb{R}^2$ with trivial bracket, $G = T^2 = \mathbb{R}/\mathbb{Z} \times \mathbb{R}/\mathbb{Z}.$ Then, if $\mathfrak{h}$ is any $1$-d subspace of irrational slope, the corresponding $H$ is a dense subset of $G$.
Lastly, yes the converse does hold. Given a Lie group $G$ and closed subgroup $H\subseteq G$, one always has $\mathfrak{h}\subseteq \mathfrak{g}$. If $\mathfrak{g}$ is defined to be, say, the set of left invariant vector fields on $G$ with vector Lie bracket, then one must show that any left $H$-invariant vector field on $H\subseteq G$ can be extended to a left $G$-invariant vector field on $G$. Then the key thing to prove is that left invariant vector fields are in 1-1 correspondence with elements of $T_e G$.