The real line can be embedded in non compact manifolds. Let $M$ be a non-compact manifold. Then $\exists$ an embedding $f:\mathbb{R}\to M$.
My attempt: I am trying to show that there is a complete smooth nonvanishing vector field on the manifold $M$ whose integral curve is not a closed curve. I took compact exhaustion and tried defining a vector field, but I seem to run into trouble.
EDIT: Embedding means: Injective immersion which is a proper map.
 A: The only proof I can think of for this ends up having a lot of moving parts, but here's a rough sketch:

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*Choose an exhaustion of $M$ by compact submanifolds with boundary $C_0\subset C_1\subset C_2,\cdots\subset M$ (such an exhaustion can be constructed from a Morse exhaustion function).

*Let $A_i=C_i\setminus\operatorname{int}(C_{i-1})$, and let $A_{i,1}\cdots,A_{i,N_i}$ be the connected components of $A_i$. (This effectively partitions $M$ into compact, connected submanifolds glued together along their common boundaries.)

*Consider $A_{i_1,j_1}$ and $A_{i_2,j_2}$ to be adjacent if they share a boundary component. This allows one to consider the $A_{i,j}$ to be the vertices of a graph.

*Show this graph is infinite and connected, and that all vertices have finite degree, implying it contains a ray $A_{i_0,j_0},A_{i_1,j_1},A_{i_2,j_2},\cdots$.

*Choose and $x_0\in\operatorname{int}(A_{i_0,j_0})$ and $x_k\in\partial A_{i_{k-1},j_{k-1}}\cap\partial A_{i_k,j_k}$.

*Construct embedded paths $\gamma_k:[0,1]\to A_{i_k,j_k}$ with $\gamma_k(0)=x_k$, $\gamma_k(1)=x_{k+1}$, and $\gamma_k((0,1))\subset\operatorname{int}(A_{i_k,j_k})$, chosen so that $\gamma_k$ and $\gamma_{k+1}$ can be smoothly concatenated.

*Show the path $\gamma:\mathbb{R}_{\ge 0}\to M$ defined by $\gamma(t)=\gamma_k(t-k)$ for $k\le t< k+1$ is a proper embedding of the ray $\mathbb{R}_{\ge 0}$.

*Construct a properly embedded  closed tubular neighborhood around $\gamma$.

*Properly embed $\mathbb{R}$ in this neighborhood.

