If $x\in {\rm span}(S\setminus \{x\})$, then ${\rm span}(S\setminus \{x\})={\rm span}S$. Let $S$ be a set of vectors from the vector space $V$. Let ${\rm span} S$ denote the set of all linear combinations of subsets of $S$. It is understood that $S$ can be finite or infinite, depending on how "big" $V$ is. I tried to prove the following statement
If there exists $x\in S$ such that $x\in {\rm span}(S\setminus \{x\})$, then ${\rm span}(S\setminus \{x\})={\rm span}S$.
The $\subseteq$ part is clear. To prove $\supseteq$, I did this way: First I pick a finite subset $X:=\{x_1,\dots,x_n\}$ from $S$, and assume that there exists $k$ such that $x_k\in {\rm span}(X\setminus \{x_k\})$. Then it can be expressed as a linear combination of $\{x_1,\dots,x_{k-1},x_{k+1},x_n\}$. Given $v\in {\rm span}X$, the vector $v$ can be expressed as a linear combination of $X$. The term $x_k$ is replaced by the other expression, showing that $v\in {\rm span}(X\setminus \{x_k\})$. Thus, I proved that ${\rm span}X\subseteq {\rm span}(X\setminus \{x_k\})$. 
The question is, how does it imply that ${\rm span}S\subseteq {\rm span}(S\setminus \{x_k\})$, if $S$ were infinite? The reason I ask is: what if one of the elements in ${\rm span}S$ doesn't belong to ${\rm span}(S\setminus \{x_k\})$, or is there a flaw in this thinking?

As you said, it is clear that the span of $S\smallsetminus x$ is contained in that of $S$. Suppose now that we can write $x = \sum \lambda_i v_i$ where none of the $v_i$ are $x$, and take $z\in \langle S\rangle$. If $z$ is a linear combination of vectors of $S$ where $x$ does not appear, then $z$ is certainly in $\langle S\smallsetminus x\rangle$. If, on the other hand, $z = \rho x + z'$ where $z'$ is in $\langle S\smallsetminus x\rangle$, it suffices to substitute $x = \sum \lambda_i v_i$ in this last equality to obtain $z$ as in the previous case. Note that, because linear combinations of vectors are always finite, it is irrelevant whether we're dealing with finite or infinite dimensional ambient spaces. 
