Prove that every subspace of a totally bounded metric space is totally bounded. Let $(Y, d)$ be a subspace of the metric space $(X, d_1)$ with the induced metric. If $(X, d_1)$ is totally bounded, then $(Y, d)$ is totally bounded; that is, every subspace of a totally bounded metric space is totally bounded.
My attempt :  If $(X, d_1)$ is totally bounded that means  $$X \subseteq \cup_{i=1}^{n} B(x_i , \epsilon)\tag1$$
If $Y$ is subspace then $Y= X \cap Y$
To prove subspace of a totally bounded metric space is totally bounded.
From $1$ we have $X \cap Y \subseteq \cup_{i=1}^{n} B(x_i , \epsilon) \cap Y$
so $Y  \subseteq \cup_{i=1}^{n} B(x_i , \epsilon) \cap Y$
Therefore every subspace of a totally bounded metric space is totally bounded.
Is my proof is valid or not ?
 A: To be fair, your proof (without considering the logical clarity you were justly told about in the comments) does imply that $Y$ is totally bounded, since you got for $Y$ a finite cover of sets with diameter less or equal to $\epsilon$. But that's an equivalence definition, as the usual definition for a totally bounded metric space would be: for every $\epsilon>0$ there is a finite cover of the space by open balls with radii equal to $\epsilon$; so, if you didn't had the equivalence at your disposal before this result, it would't make much sense to use it for the proof. I'll leave here a proof using the usual definition:
Given $\epsilon>0$, take $\epsilon_0=\dfrac{\epsilon}{2}$. Since $X$ is totally bounded, there has to be a finite cover of open balls with radii equal to $\epsilon_0$. So there exist a finite amount of $x_i\in X$ (say there are $n$) such that $X\subseteq \cup_{i=1}^n B(x_i;\epsilon_0)$. Now, $Y\subseteq X\subseteq \cup_{i=1}^n B(x_i;\epsilon_0)$, so those open balls are covering $Y$ too.
Consider the $x_i$ such that $B(x_i;\epsilon_0)$ contain at least one element of $Y$ and name them $x_{i_k}$. We can get rid of the open balls that doesn't contain any element of $Y$, and keep the ones that do (say there are $l\le n$), so $Y\subseteq\cup_{k=1}^l B(x_{i_k};\epsilon_0)$. Take one element that is in $Y$ from each $B(x_{i_k};\epsilon_0)$, we can name them $y_k$ (it could be $x_{i_k}$ itself).
Remember $B(x_{i_k};\epsilon_0)=\{x\in X:d_1(x,x_{i_k})<\epsilon_0\}$, so let's see what happens if we consider $B(y_k,\epsilon)$. If $x\in B(x_{i_k},\epsilon_0)$ then $d_1(x,x_{i_k})<\epsilon_0=\dfrac{\epsilon}{2}$, so $d(x,y_k)\le d(x,x_{i_k})+d(x_{i_k},y_k)<\epsilon_0+\epsilon_0=\epsilon$, since $y_k\in B(x_{i_k},\epsilon_0)$. Therefore $x\in B(y_k,\epsilon)$, so $B(x_{i_k},\epsilon_0)\subseteq B(y_k,\epsilon)$ for every $k$.
We are nearly done, but first, to be completely rigorous, we have to get rid of the elements of $B(y_k,\epsilon)$ that are not in $Y$, since we want a cover for $Y$ in $Y$. We also want that cover to be of open balls for the induced metric in $Y$, so we can simply consider $B'(y_k,\epsilon)=\{y\in Y:d(y,y_k)<\epsilon\}$. This doesn't ruin the contentions we are interested in, since the analogous open balls $B'(x_{i_k};\epsilon_0)=\{y\in Y:d_1(y,x_{i_k})<\epsilon_0\}$ just get rid of the elements of the elements that weren't in $Y$, so $Y\subseteq \cup_{k=1}^l B'(x_{i_k};\epsilon_0)$ and $B'(x_{i_k};\epsilon_0)\subseteq B'(y_k,\epsilon)$ for every $k$. Therefore $Y\subseteq \cup_{k=1}^l B'(x_{i_k};\epsilon_0)\subseteq\cup_{k=1}^l B'(y_k,\epsilon)$, so $Y$ is covered by a finite amount of open balls (open for its metric) with radii $\epsilon$. Since $\epsilon$ was arbitrary, $Y$ is totally bounded.
