Let $X$ a space and $Y \subseteq X$ a subspace. Then $Y$ is connected if $(Y,\tau_Y)$ is connected, where $\tau_Y$ is the subspace topology inherited from $X$.
My question:
Let $X$ a topological space and $V \subseteq U \subseteq X$. If $V$ is connected (that´s mean $(V,\tau_V)$ is connected) then ($V,\tau')$ is connected? where $\tau'$ is the subspace topology inherited from $U$.
My "proof":
If $M \in \tau'$ then $M= W \cap V$ where $W$ is open in $\tau_U$, so $W=W_1 \cap U$ where $W_1$ is open in $X$. Then $M= W_1 \cap U \cap V = W_1 \cap V$. So $M \in \tau_V$.
That is $\tau' \subseteq \tau_V$. How $(V,\tau_V)$ is connected then $(V,\tau')$ is connected.
We can understand it as follows: if $V$ is connected and $V \subseteq U$ then $V$ will also be connected, seeing $U$ as the total space?.
There is a very general fact called "the transitive law of initial topologies" of which you have redicovered a special case:
If $X$ is a space and $A \subseteq B \subseteq X$, then the subspace topology of $A$ w.r.t. $\tau_B$ (the subspace topology that $B$ inherits from $X$) is identical to $\tau_A$ (the subspace topology that $A$ directly inherits from $X$). So e.g. for connectedness of $A$ it suffices to look at whether $(A, \tau_A)$ is connected as a space in its own right.
So "yes" to your final question.