Let $E, F, \subseteq V$ for a K-vector space V. If $E \subseteq F$, then $\left \subseteq \left$ I tried the following:
Let $x \in \left<E\right>$. Therefore, x is a lineal combination of elements of E ($x = \sum_{i=1}^n a_ie_i,\ a_i \in K,\ e_i \in E$). Now, since $E \subseteq F$, then $x = \sum_{i=1}^n a_ie_i,\ a_i \in K,\ e_i \in F$. Therefore, $x \in \left<F\right>$ and $\left<E\right> \subseteq \left<F\right>$.
I'm unsure whether my answer is correct.
 A: There are lots of confusions in your attempt.

*

*You wanted to show $\langle E\rangle\subseteq\langle F\rangle$ but you ended up with $\langle E\rangle\subseteq\langle V\rangle$.

*"$\langle E\rangle$ is a set of elements $\{e_1,\cdots,e_n\}$ every vector v... $\forall a_i\in K$" does not make sense.
By definition,  $\langle E\rangle$ means
$$
\langle E\rangle=\left\{
\sum _{i=1}^{k}\lambda _{i}v_{i}\;:
\;k\in \mathbb {N} ,v_{i}\in E,\lambda _{i}\in K
\right\}.\tag{1}
$$
namely all the linear combinations of elements in $E$.

It follows immediately from (1) that $\langle E\rangle\subseteq\langle F\rangle$: because $E$ is a subset of $F$, and thus any linear combinations of elements in $E$ must also be a combination of elements in $F$.
A: It is even simpler if you use that $\langle E\rangle$ is the smallest subspace of $V$ containing $E$. So, assume that $E\subseteq F$. Since $F\subseteq \langle F\rangle$, you know that $E\subseteq \langle F\rangle$.  Since $\langle F\rangle$ is a subspace of $V$, the first sentence applies to yield $\langle E\rangle\subseteq \langle F\rangle$.
