connected 2D manifold contained in a 2D affine plane Let $M\subset\mathbb{R}^4$ be a connected manifold of dimension $2$.
Suppose that $\forall x,y\in M: T_xM=T_yM$.
Prove that $M$ is contained in a two dimensional affine plane, i.e. there exists a two dimensional linear space $L\subset \mathbb{R}^4$ and a vector $v\in\mathbb{R}^4$ such that $M\subset \{v+x|x\in L\}$.
Note: The fact that $M$ is connected implies that there is no such decomposition $M=M_1\cup M_2$ such that $M_1\cap M_2=\emptyset$ and $M_1,M_2$ are relatively open in $M$.
Can't we just fix some $x_0\in M$ and say that since $\forall y:y\in T_yM$ and $\forall y:T_{x_0}M=T_yM$ we have that $\forall y:y\in T_{x_0}M$, so finally $M\subseteq T_{x_0}M$ (and $T_{x_0}M$ is a linear space of dimension $2$).
I suspect that what I said is wrong as I didn't use the connectivity. 
 A: It seems to me there is a problem with the OP's argument, since, in general,
$y \notin T_yM. \tag 1$
This may be seen with the aid of the following counterexample:  let
$y = (1, 0, 0, 0) \in \Bbb R^4, \tag 2$
and let $M$ be the set
$M = \{ y  + s(0, 1, 0, 0) + t(0, 0, 1, 0)  = (1, s, t, 0)  \mid s, t \in \Bbb R \} \subset \Bbb R^4; \tag 3$
then $M$ is an affine two-plane lying in $\Bbb R^4$, and we have
$T_yM = \{s(0, 1, 0, 0) + t(0, 0, 1, 0) = (0, s, t, 0)\mid s, t \in \Bbb R \}; \tag 4$
but
$y = (1, 0, 0, 0) \ne (0, s, t, 0) \tag 5$
for any $s, t \in \Bbb R$.  Thus (1) holds for this example.
The requisite assertion may be proved as follows; note that the argument relies on the connectedness of $M$:
Since $M$ is a connected manifold, it is also path-connected; this is a standard and well-known result in the topology of manifolds.
For $z, y \in M$, let 
$\gamma: [0, 1] \to M \tag 6$
be a differentiable path in $M$ such that
$\gamma(0) = y, \; \gamma(1) = z; \tag 7$
then
$z - y = \gamma(1) - \gamma(0) = \displaystyle \int_0^1 \dot \gamma(s)\;ds; \tag 8$
we next observe that
$\dot \gamma(t) \in T_{\gamma(t)}M = T_yM \tag 9$
since by our hypothesis $T_x = T_y$ for all $x, y \in M$.  From this we may affirm that
$\displaystyle \int_0^1 \dot \gamma(s)\;ds \in T_yM, \tag{10}$
which may be seen as follows:  clearly, 
$u \in T_yM \Longleftrightarrow \forall w \in (T_yM)^\bot, \; w \cdot u = 0; \tag{11}$
by (9), 
$\dot \gamma(t) \in T_y , \tag{12}$
whence
$ \displaystyle w \cdot \int_0^1 \dot \gamma(s)\;ds =  \int_0^1 w \cdot \dot \gamma(s)\;ds = \int_0^1 0 \; ds = 0, \forall w \in (T_yM)^\bot; \tag{13}$
thus, by (11), we may infer (10); writing (8) in the form
$z = y + \displaystyle \int_0^1 \dot \gamma(s) \; = y + u, \tag{14}$
where
$u = \displaystyle \int_0^1 \dot \gamma(s) \; ds \in T_yM, \tag{15}$
we thus see that
$M \subset \{ y + u \mid u \in T_yM \}, \tag{16}$
which is in the requisite format.
It remains to validate our assertion that we we may take $\gamma(t)$ to be a differentiable path 'twixt $y$ and any $z \in M$;  I argue this as follows:  that $M$ is a differentiable manifold of class at least $C^1$ is implicit in the statement of the problem, since we are given the existence of tangent spaces $T_yM$ for every $y \in M$; thus choosing local systems of $C^1$ coordinates for $M$, we may use it to explicitly specify $\gamma(t)$; such a construction is often used in differential topology and need not be re-iterated in detail here.
Finally, we should observe that there is nothing essential about choosing $M \subset \Bbb R^4$ to be a two-manifold; we could just as well take $\dim M = m$ in $\Bbb R^n$ with $n \ge m$ and the entire argument goes through.
