Statement: Given 2 two-dimensional manifolds $\emptyset \neq N \subseteq M \subseteq \mathbb{R}^3 $, if $N$ is compact and $M$ is path-connected, then $N = M$.
Proof: We know that there is at least one point $p\in N$ and we assume that there exists another point $q \notin N, q \in M$. From path-connectivity we get that there exists a path $\gamma(t) : [0,1] \rightarrow M$, such that $\gamma(0) = p$ and $\gamma(1) = q$.
Label $t_0$ that maximum $t \in [0,1]$ such that $\gamma(t) \in N$. This exists an is well defined because $N$ is compact. We label $\gamma(t_0) = x_0$.
Claim: $T_{x_0}M = T_{x_0}N$. From $N \subseteq M$ we get $T_{x_0}N \subseteq T_{x_0}M$. This, coupled with the fact that they are both 2-dimensional linear spaces in $\mathbb{R}^3$ and therefore they have to be equal.
** I now want to say that because of this, I can "continue going along with the path" and get closer to $q$ while staying on $N$, and therefore $t_0$ isn't maximal, a contradiction.
I'm not really sure how to formally express this, I know that intuitively, the the tangent space is an "up close" look at the manifold and that I can go along the tangent space and stay on the manifold, but I'm not sure how to formally say this.
Any help would be welcome!
Your proof so far is good, but there is no need to pass to the tangent space.
Sketch: The point $x_0$ has an open neighborhood $U\subseteq N$ homeomorphic to an open ball in $\mathbb{R}^n$, so there is an induced open ball $B\subseteq[0,1]$ at $t_0\in B\subseteq[0,1]$ such that the path $\gamma(B)\subseteq U$. This then implies that $t_0=1$ lest we contradict the definition of $t_0$.
NB: The above sketch of a proof assumes that $N$ is a manifold without boundary. If $N$ has boundary, then the statement is false (e.g. look at a closed disk living in the sphere).